Thể tích khối nón đã cho là:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaiodaaaGaeqiWdaNaamOuamaaCaaa % leqabaGaaGOmaaaakiaadIgacqGH9aqpdaWcaaqaaiaaigdaaeaaca % aIZaaaaiabec8aWjaac6cacaaIYaGaamyyaiaac6cacaWGHbWaaWba % aSqabeaacaaIYaaaaOGaeyypa0ZaaSaaaeaacaaIYaGaeqiWdaNaam % yyamaaCaaaleqabaGaaG4maaaaaOqaaiaaiodaaaaaaa!4D21! V = \frac{1}{3}\pi {R^2}h = \frac{1}{3}\pi .2a.{a^2} = \frac{{2\pi {a^3}}}{3}\)

Ta có thể tích của khối chóp đã cho là:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGtbGaaiOlaiaadgeacaWGcbGaam4qaiaadseaaeqaaOGa % eyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacaWGtbGaamyqaiaac6 % cacaWGtbWaaSbaaSqaaiaadgeacaWGcbGaam4qaiaadseaaeqaaOGa % eyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacaGGUaGaamyyaiaac6 % cacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaeyypa0ZaaSaaaeaacaWG % HbWaaWbaaSqabeaacaaIZaaaaaGcbaGaaG4maaaaaaa!4F15! {V_{S.ABCD}} = \frac{1}{3}SA.{S_{ABCD}} = \frac{1}{3}.a.{a^2} = \frac{{{a^3}}}{3}\)

Đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaai % OoamaalaaabaGaamiEaiabgkHiTiaaigdaaeaacaaIXaaaaiabg2da % 9maalaaabaGaamyEaiabgUcaRiaaiodaaeaacaaIYaaaaiabg2da9m % aalaaabaGaamOEaiabgkHiTiaaiodaaeaacqGHsislcaaI1aaaaaaa % !4562! \Delta :\frac{{x - 1}}{1} = \frac{{y + 3}}{2} = \frac{{z - 3}}{{ - 5}}\) nhận vecto ( 1;2;-5) làm 1 VTCP

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaSaaaeaacaWGHbaabaGa % amOyamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpciGGSbGaai4Bai % aacEgadaWgaaWcbaGaaGOmaaqabaGccaWGHbGaeyOeI0IaciiBaiaa % c+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaamOyamaaCaaaleqaba % GaaGOmaaaakiabg2da9iGacYgacaGGVbGaai4zamaaBaaaleaacaaI % YaaabeaakiaadggacqGHsislcaaIYaGaciiBaiaac+gacaGGNbWaaS % baaSqaaiaaikdaaeqaaOGaamOyaaaa!54BD! {\log _2}\frac{a}{{{b^2}}} = {\log _2}a - {\log _2}{b^2} = {\log _2}a - 2{\log _2}b\)

Mặt phẳng trung trực \(\alpha\) của đoạn thẳng AB nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGbbGaamOqaaGaay51Gaaaaa!3935! \overrightarrow {AB} \) làm một VTPT.

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGbbGaamOqaaGaay51GaGaeyypa0ZaaeWaaeaacaaIYaGaai4oaiaa % isdacaGG7aGaeyOeI0IaaGOmaaGaayjkaiaawMcaaiabg2da9iaaik % dadaqadaqaaiaaigdacaGG7aGaaGOmaiaacUdacqGHsislcaaIXaaa % caGLOaGaayzkaaGaai4laiaac+cadaqadaqaaiaaigdacaGG7aGaaG % OmaiaacUdacqGHsislcaaIXaaacaGLOaGaayzkaaaaaa!4FD9! \overrightarrow {AB} = \left( {2;4; - 2} \right) = 2\left( {1;2; - 1} \right)//\left( {1;2; - 1} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aae % WaaeaacqaHXoqyaiaawIcacaGLPaaaaaa!3B79! \Rightarrow \left( \alpha \right)\) nhận vecto ( 1;2;-1) làm 1 VTPT

Gọi q là công bội của CSN đã cho, ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaaIXaaabeaakiabg2da9iaaigdacaGG7aGaamyDamaaBaaa % leaacaaIYaaabeaakiabg2da9iabgkHiTiaaikdacqGHshI3caWGXb % Gaeyypa0ZaaSaaaeaacaWG1bWaaSbaaSqaaiaaikdaaeqaaaGcbaGa % amyDamaaBaaaleaacaaIXaaabeaaaaGccqGH9aqpdaWcaaqaaiabgk % HiTiaaikdaaeaacaaIXaaaaiabg2da9iabgkHiTiaaikdaaaa!4D63! {u_1} = 1;{u_2} = - 2 \Rightarrow q = \frac{{{u_2}}}{{{u_1}}} = \frac{{ - 2}}{1} = - 2\)

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yDamaaBaaaleaacaaIYaGaaGimaiaaigdacaaI5aaabeaakiabg2da % 9iaadwhadaWgaaWcbaGaaGymaaqabaGccaGGUaGaamyCamaaCaaale % qabaGaaGOmaiaaicdacaaIXaGaaGioaaaakiabg2da9iaaigdacaGG % UaWaaeWaaeaacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqabe % aacaaIYaGaaGimaiaaigdacaaI4aaaaOGaeyypa0JaaGOmamaaCaaa % leqabaGaaGOmaiaaicdacaaIXaGaaGioaaaaaaa!51E9! \Rightarrow {u_{2019}} = {u_1}.{q^{2018}} = 1.{\left( { - 2} \right)^{2018}} = {2^{2018}}\)

Dựa vào đồ thị hàm số ta thấy đồ thị hàm số có dạng 1 parabol có đỉnh là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaai4oaiabgkHiTiaaikdaaiaawIcacaGLPaaacqGHshI3aaa!3CFC! \left( {0; - 2} \right) \Rightarrow \) loại đáp án A, D.

Đồ thị hàm số đi qua các điểm (1;0) và (-1;0), thay tọa độ các điểm này vào công thức hàm số ở đáp án B và C thấy chỉ có đáp án B thỏa mãn.

Có 1 điểm cực trị có tọa độ là (0;-2)

Gọi R là bán kính mặt cầu cần tìm  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % Ouaiabg2da9iaadsgadaqadaqaaiaadMeacaGG7aWaaeWaaeaacqaH % XoqyaiaawIcacaGLPaaaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaam % aaemaabaGaaGymaiabgkHiTiaaikdacaGGUaGaaGOmaiabgUcaRiaa % ikdacaGGUaGaaGynaiabgUcaRiaaikdaaiaawEa7caGLiWoaaeaada % GcaaqaaiaaigdacqGHRaWkdaqadaqaaiabgkHiTiaaikdaaiaawIca % caGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaWaaWbaaS % qabeaacaaIYaaaaaqabaaaaOGaeyypa0ZaaSaaaeaacaaI5aaabaGa % aG4maaaacqGH9aqpcaaIZaaaaa!5AC8! \Rightarrow R = d\left( {I;\left( \alpha \right)} \right) = \frac{{\left| {1 - 2.2 + 2.5 + 2} \right|}}{{\sqrt {1 + {{\left( { - 2} \right)}^2} + {2^2}} }} = \frac{9}{3} = 3\)

Vậy mặt cầu tâm I và tiếp xúc với \((\alpha)\) có phương trình là:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaakiabgUcaRmaabmaabaGaamyEaiabgkHiTiaaikdaaiaawIcaca % GLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkdaqadaqaaiaadQha % cqGHsislcaaI1aaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaO % Gaeyypa0JaaGyoaaaa!48EC! {\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 5} \right)^2} = 9\)

Dựa vào đồ thị hàm số ta thấy, trên đoạn [-3;3], hàm số  có 3 điểm cực trị là ( -1;1);(1; -3) ; (2;-3)

Sử dụng các tính chất của tích phân: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WadaqaaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGHXcqS % caWGNbWaaeWaaeaacaWG4baacaGLOaGaayzkaaaacaGLBbGaayzxaa % aaleaacaWGHbaabaGaamOyaaqdcqGHRiI8aOGaamizaiaadIhacqGH % 9aqpdaWdXbqaaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaaca % WGKbGaamiEaaWcbaGaamyyaaqaaiaadkgaa0Gaey4kIipakiabggla % XoaapehabaGaam4zamaabmaabaGaamiEaaGaayjkaiaawMcaaiaads % gacaWG4baaleaacaWGHbaabaGaamOyaaqdcqGHRiI8aaaa!5CF6! \int\limits_a^b {\left[ {f\left( x \right) \pm g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)dx} \pm \int\limits_a^b {g\left( x \right)dx} \)

Dựa vào đồ thị hàm số ta thấy hàm số đồng biến trên các khoảng (-3;-1) và (1;2)

Hàm số nghịch biến trên các khoảng: (-1;1) và (2;3)

Ta có:

+) Đáp án A: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIYaWaaOaaaeaacaaIZaGaamiEaiabgkHiTiaaikdaaSqabaGccqGH % RaWkcaWGdbaacaGLOaGaayzkaaGaai4jaiabg2da9maalaaabaGaaG % Omaiaac6cacaaIZaaabaGaaGOmamaakaaabaGaaG4maiaadIhacqGH % sislcaaIYaaaleqaaaaakiabg2da9maalaaabaGaaG4maaqaamaaka % aabaGaaG4maiaadIhacqGHsislcaaIYaaaleqaaaaakiabgcMi5oaa % laaabaGaaGymaaqaamaakaaabaGaaG4maiaadIhacqGHsislcaaIYa % aaleqaaaaakiabgkDiEdaa!536D! \left( {2\sqrt {3x - 2} + C} \right)' = \frac{{2.3}}{{2\sqrt {3x - 2} }} = \frac{3}{{\sqrt {3x - 2} }} \ne \frac{1}{{\sqrt {3x - 2} }} \Rightarrow \) đáp án A sai.

+) Đáp án B:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaada % WcaaqaaiaaikdaaeaacaaIZaaaamaakaaabaGaaG4maiaadIhacqGH % sislcaaIYaaaleqaaOGaey4kaSIaam4qaaGaayjkaiaawMcaaiaacE % cacqGH9aqpdaWcaaqaaiaaikdacaGGUaGaaG4maaqaaiaaiodacaGG % UaGaaGOmamaakaaabaGaaG4maiaadIhacqGHsislcaaIYaaaleqaaa % aakiabg2da9maalaaabaGaaGymaaqaamaakaaabaGaaG4maiaadIha % cqGHsislcaaIYaaaleqaaaaakiabgkDiEdaa!4F8D! \left( {\frac{2}{3}\sqrt {3x - 2} + C} \right)' = \frac{{2.3}}{{3.2\sqrt {3x - 2} }} = \frac{1}{{\sqrt {3x - 2} }} \Rightarrow \)  đáp án B đúng.

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGyoamaaCa % aaleqabaGaamiEaiabgUcaRiaaigdaaaGccqGHsislcaaIZaWaaWba % aSqabeaacaWG4bGaey4kaSIaaGymaaaakiabgkHiTiaaiodacaaIWa % Gaeyypa0JaaGimaiabgsDiBlaaiMdacaGGUaGaaGyoamaaCaaaleqa % baGaamiEaaaakiabgkHiTiaaiodacaGGUaGaaG4mamaaCaaaleqaba % GaamiEaaaakiabgkHiTiaaiodacaaIWaGaeyypa0JaaGimaiabgsDi % BlaaiodacaGGUaWaaeWaaeaacaaIZaWaaWbaaSqabeaacaWG4baaaa % GccaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG4m % amaaCaaaleqabaGaamiEaaaakiabgkHiTiaaigdacaaIWaGaeyypa0 % JaaGimamaabmaabaGaaiOkaaGaayjkaiaawMcaaaaa!61CF! {9^{x + 1}} - {3^{x + 1}} - 30 = 0 \Leftrightarrow {9.9^x} - {3.3^x} - 30 = 0 \Leftrightarrow 3.{\left( {{3^x}} \right)^2} - {3^x} - 10 = 0\left( * \right)\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4mamaaCa % aaleqabaGaamiEaaaakiabg2da9iaadshaaaa!39E4! {3^x} = t\) ta có phương trình (*)  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % 4maiaadshadaahaaWcbeqaaiaaikdaaaGccqGHsislcaWG0bGaeyOe % I0IaaGymaiaaicdacqGH9aqpcaaIWaaaaa!4101! \Leftrightarrow 3{t^2} - t - 10 = 0\)

Cách 1: Gọi số cần tìm có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaamOyaiaadogaaaaaaa!38BA! \overline {abc} \) là số cần lập.

Chọn 3 số a,b,c bất kì trong 9 số ta có: \(A^3_9\) cách chọn.

Cách 2: Gọi số cần tìm có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaamOyaiaadogaaaaaaa!38BA! \overline {abc} \) là số cần lập.

Khi đó a có 9 cách chọn.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOyaiabgc % Mi5kaadggacqGHshI3caWGIbaaaa!3CCC! b \ne a \Rightarrow b\) có 8 cách chọn.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yaiabgc % Mi5kaadggacaGGSaGaam4yaiabgcMi5kaadkgacqGHshI3caWGJbaa % aa!4114! c \ne a,c \ne b \Rightarrow c\)có 7 cách chọn.

\(\Rightarrow\) có 9.8.7 = \(A^3_9\) = 504 cách chọn.

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEaiabg2 % da9iabgkHiTiaaikdacqGHRaWkcaWGPbGaeyO0H49aa0aaaeaacaWG % 6baaaiabg2da9iabgkHiTiaaikdacqGHsislcaWGPbGaeyO0H4Taam % OtamaabmaabaGaeyOeI0IaaGOmaiaacUdacqGHsislcaaIXaaacaGL % OaGaayzkaaaaaa!4C32! z = - 2 + i \Rightarrow \overline z = - 2 - i \Rightarrow N\left( { - 2; - 1} \right)\) là điểm biểu diễn số phức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WG6baaaaaa!3704! \overline z \) 

Ta có: \(\Delta_1\)  có VTCP là:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1bWaaSbaaSqaaiaaigdaaeqaaaGccaGLxdcacqGH9aqpdaqadaqa % aiabgkHiTiaaikdacaGG7aGaaGymaiaacUdacaaIYaaacaGLOaGaay % zkaaaaaa!40C0!$ \overrightarrow {{u_1}} = \left( { - 2;1;2} \right)\) , \(\Delta_2\) có VTCP là:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1bWaaSbaaSqaaiaaikdaaeqaaaGccaGLxdcacqGH9aqpdaqadaqa % aiaaigdacaGG7aGaaGymaiaacUdacqGHsislcaaI0aaacaGLOaGaay % zkaaaaaa!40C2! \overrightarrow {{u_2}} = \left( {1;1; - 4} \right)\)

Gọi \(\alpha\) là góc giữa hai đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaS % baaSqaaiaaigdaaeqaaOGaaiilaiabfs5aenaaBaaaleaacaaIYaaa % beaaaaa!3B49! {\Delta _1},{\Delta _2}\) ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbGaeqySdeMaeyypa0ZaaSaaaeaadaabdaqaamaaFiaabaGa % amyDamaaBaaaleaacaaIXaaabeaaaOGaay51GaGaaiOlamaaFiaaba % GaamyDamaaBaaaleaacaaIYaaabeaaaOGaay51GaaacaGLhWUaayjc % SdaabaWaaqWaaeaadaWhcaqaaiaadwhadaWgaaWcbaGaaGymaaqaba % aakiaawEniaaGaay5bSlaawIa7aiaac6cadaabdaqaamaaFiaabaGa % amyDamaaBaaaleaacaaIYaaabeaaaOGaay51GaaacaGLhWUaayjcSd % aaaiabg2da9maalaaabaWaaqWaaeaacqGHsislcaaIYaGaaiOlaiaa % igdacqGHRaWkcaaIXaGaaiOlaiaaigdacqGHRaWkcaaIYaGaaiOlam % aabmaabaGaeyOeI0IaaGinaaGaayjkaiaawMcaaaGaay5bSlaawIa7 % aaqaamaakaaabaWaaeWaaeaacqGHsislcaaIYaaacaGLOaGaayzkaa % WaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaiabgUcaRiaaikda % daahaaWcbeqaaiaaikdaaaaabeaakiaac6cadaGcaaqaaiaaigdacq % GHRaWkcaaIXaGaey4kaSYaaeWaaeaacqGHsislcaaI0aaacaGLOaGa % ayzkaaWaaWbaaSqabeaacaaIYaaaaaqabaaaaOGaeyypa0ZaaSaaae % aacaaI5aaabaGaaG4maiaac6cacaaIZaWaaOaaaeaacaaIYaaaleqa % aaaakiabg2da9maalaaabaWaaOaaaeaacaaIYaaaleqaaaGcbaGaaG % Omaaaaaaa!7CBE! \cos \alpha = \frac{{\left| {\overrightarrow {{u_1}} .\overrightarrow {{u_2}} } \right|}}{{\left| {\overrightarrow {{u_1}} } \right|.\left| {\overrightarrow {{u_2}} } \right|}} = \frac{{\left| { - 2.1 + 1.1 + 2.\left( { - 4} \right)} \right|}}{{\sqrt {{{\left( { - 2} \right)}^2} + 1 + {2^2}} .\sqrt {1 + 1 + {{\left( { - 4} \right)}^2}} }} = \frac{9}{{3.3\sqrt 2 }} = \frac{{\sqrt 2 }}{2}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taeq % ySdeMaeyypa0JaaGinaiaaiwdadaahaaWcbeqaaiaaicdaaaaaaa!3D5A! \Rightarrow \alpha = {45^0}\)

Gọi số phức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEaiabg2 % da9iaadggacqGHRaWkcaWGIbGaamyAaiaacYcadaqadaqaaiaadgga % caGGSaGaamOyaiabgIGiolabl2riHcGaayjkaiaawMcaaaaa!4340! z = a + bi,(a,b \in R)\). Khi đó ta có:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WG6baaaiabg2da9iaadggacqGHsislcaWGIbGaamyAaaaa!3BB2! \overline z = a - bi\)

Khi đó ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG6b % Gaey4kaSIaaGOmamaanaaabaGaamOEaaaacqGH9aqpcaaI2aGaey4k % aSIaaGOmaiaadMgacqGHuhY2caWGHbGaey4kaSIaamOyaiaadMgacq % GHRaWkcaaIYaWaaeWaaeaacaWGHbGaeyOeI0IaamOyaiaadMgaaiaa % wIcacaGLPaaacqGH9aqpcaaI2aGaey4kaSIaaGOmaiaadMgaaeaacq % GHuhY2caaIZaGaamyyaiabgkHiTiaadkgacaWGPbGaeyypa0JaaGOn % aiabgUcaRiaaikdacaWGPbGaeyi1HS9aaiqaaqaabeqaaiaaiodaca % WGHbGaeyypa0JaaGOnaaqaaiabgkHiTiaadkgacqGH9aqpcaaIYaaa % aiaawUhaaiabgsDiBpaaceaaeaqabeaacaWGHbGaeyypa0JaaGOmaa % qaaiaadkgacqGH9aqpcqGHsislcaaIYaaaaiaawUhaaiabgkDiElaa % dQhacqGH9aqpcaaIYaGaeyOeI0IaaGOmaiaadMgaaaaa!7613! \begin{array}{l} z + 2\overline z = 6 + 2i \Leftrightarrow a + bi + 2\left( {a - bi} \right) = 6 + 2i\\ \Leftrightarrow 3a - bi = 6 + 2i \Leftrightarrow \left\{ \begin{array}{l} 3a = 6\\ - b = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 2\\ b = - 2 \end{array} \right. \Rightarrow z = 2 - 2i \end{array}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ytamaabmaabaGaaGOmaiaacUdacqGHsislcaaIYaaacaGLOaGaayzk % aaaaaa!3DD0! \Rightarrow M\left( {2; - 2} \right)\) là điểm biểu diễn số phức z.

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaWcaaqaaiaadIhacqGHsislcaaIYaaabaGaeyOeI0IaaGymaaaa % cqGH9aqpdaWcaaqaaiaadMhacqGHsislcaaIXaaabaGaaGOmaaaacq % GH9aqpdaWcaaqaaiaadQhaaeaacaaIYaaaaiabgkDiElaadsgacaGG % 6aWaaiqaaqaabeqaaiaadIhacqGH9aqpcaaIYaGaeyOeI0IaamiDaa % qaaiaadMhacqGH9aqpcaaIXaGaey4kaSIaaGOmaiaadshaaeaacaWG % 6bGaeyypa0JaaGOmaiaadshaaaGaay5EaaGaeyO0H4Taamytamaabm % aabaGaaGOmaiabgkHiTiaadshacaGG7aGaaGymaiabgUcaRiaaikda % caWG0bGaai4oaiaaikdacaWG0baacaGLOaGaayzkaaaaaa!63FC! d:\frac{{x - 2}}{{ - 1}} = \frac{{y - 1}}{2} = \frac{z}{2} \Rightarrow d:\left\{ \begin{array}{l} x = 2 - t\\ y = 1 + 2t\\ z = 2t \end{array} \right. \Rightarrow M\left( {2 - t;1 + 2t;2t} \right)\)   là một điểm thuộc đường thẳng d.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGnb % Gaeyypa0JaamizaiabgMIihpaabmaabaGaamiuaaGaayjkaiaawMca % aiabgkDiElaaikdacqGHsislcaWG0bGaey4kaSIaaGOmamaabmaaba % GaaGymaiabgUcaRiaaikdacaWG0baacaGLOaGaayzkaaGaeyOeI0Ya % aeWaaeaacaaIYaGaamiDaaGaayjkaiaawMcaaiabgkHiTiaaiwdacq % GH9aqpcaaIWaaabaGaeyi1HSTaamiDaiabg2da9iaaigdacqGHshI3 % caWGnbWaaeWaaeaacaaIXaGaai4oaiaaiodacaGG7aGaaGOmaaGaay % jkaiaawMcaaaaaaa!5D47! \begin{array}{l} M = d \cap \left( P \right) \Rightarrow 2 - t + 2\left( {1 + 2t} \right) - \left( {2t} \right) - 5 = 0\\ \Leftrightarrow t = 1 \Rightarrow M\left( {1;3;2} \right) \end{array}\)

Điều kiện : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIZaGaamiE % aiabg6da+iaaicdaaeaacaaI5aGaeyOeI0IaamiEaiabg6da+iaaic % daaaGaay5EaaGaeyi1HS9aaiqaaqaabeqaaiaadIhadaqadaqaaiaa % dIhacqGHsislcaaIZaaacaGLOaGaayzkaaGaeyOpa4JaaGimaaqaai % aadIhacqGH8aapcaaI5aaaaiaawUhaaiabgsDiBpaaceaaeaqabeaa % daWabaabaeqabaGaamiEaiabg6da+iaaiodaaeaacaWG4bGaeyipaW % JaaGimaaaacaGLBbaaaeaacaWG4bGaeyipaWJaaGyoaaaacaGL7baa % cqGHuhY2daWabaabaeqabaGaamiEaiabgYda8iaaicdaaeaacaaIZa % GaeyipaWJaamiEaiabgYda8iaaiMdaaaGaay5waaaaaa!666B! \left\{ \begin{array}{l} {x^2} - 3x > 0\\ 9 - x > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x\left( {x - 3} \right) > 0\\ x < 9 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x > 3\\ x < 0 \end{array} \right.\\ x < 9 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x < 0\\ 3 < x < 9 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGSb % Gaai4BaiaacEgadaWgaaWcbaGaaGinaaqabaGcdaqadaqaaiaadIha % daahaaWcbeqaaiaaikdaaaGccqGHsislcaaIZaGaamiEaaGaayjkai % aawMcaaiabg6da+iGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaa % beaakmaabmaabaGaaGyoaiabgkHiTiaadIhaaiaawIcacaGLPaaacq % GHuhY2daWcaaqaaiaaigdaaeaacaaIYaaaaiGacYgacaGGVbGaai4z % amaaBaaaleaacaaIYaaabeaakmaabmaabaGaamiEamaaCaaaleqaba % GaaGOmaaaakiabgkHiTiaaiodacaWG4baacaGLOaGaayzkaaGaeyOp % a4JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaae % aacaaI5aGaeyOeI0IaamiEaaGaayjkaiaawMcaaaqaaiabgsDiBlGa % cYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam % iEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiodacaWG4baacaGL % OaGaayzkaaGaeyOpa4JaaGOmaiGacYgacaGGVbGaai4zamaaBaaale % aacaaIYaaabeaakmaabmaabaGaaGyoaiabgkHiTiaadIhaaiaawIca % caGLPaaacqGHuhY2ciGGSbGaai4BaiaacEgadaWgaaWcbaGaaGOmaa % qabaGcdaqadaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsisl % caaIZaGaamiEaaGaayjkaiaawMcaaiabg6da+iGacYgacaGGVbGaai % 4zamaaBaaaleaacaaIYaaabeaakmaabmaabaGaaGyoaiabgkHiTiaa % dIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaakeaacqGHuh % Y2caWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG4maiaadIha % cqGH+aGpcaaI4aGaaGymaiabgkHiTiaaigdacaaI4aGaamiEaiabgU % caRiaadIhadaahaaWcbeqaaiaaikdaaaaakeaacqGHuhY2caaIXaGa % aGynaiaadIhacqGH+aGpcaaI4aGaaGymaiabgsDiBlaadIhacqGH+a % GpdaWcaaqaaiaaiIdacaaIXaaabaGaaGymaiaaiwdaaaGaeyi1HSTa % amiEaiabg6da+maalaaabaGaaGOmaiaaiEdaaeaacaaI1aaaaaaaaa!B0EF! \begin{array}{l} {\log _4}\left( {{x^2} - 3x} \right) > {\log _2}\left( {9 - x} \right) \Leftrightarrow \frac{1}{2}{\log _2}\left( {{x^2} - 3x} \right) > {\log _2}\left( {9 - x} \right)\\ \Leftrightarrow {\log _2}\left( {{x^2} - 3x} \right) > 2{\log _2}\left( {9 - x} \right) \Leftrightarrow {\log _2}\left( {{x^2} - 3x} \right) > {\log _2}{\left( {9 - x} \right)^2}\\ \Leftrightarrow {x^2} - 3x > 81 - 18x + {x^2}\\ \Leftrightarrow 15x > 81 \Leftrightarrow x > \frac{{81}}{{15}} \Leftrightarrow x > \frac{{27}}{5} \end{array}\)

Kết hợp với điều kiện xác định ta có bất phương trình có tập nghiệm là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaaG4naaqaaiaaiwdaaaGaeyipaWJaamiEaiabgYda8iaaiMda % aaa!3C08! \frac{{27}}{5} < x < 9\)

Mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiolablssiIkabgkDiElaadIhacqGHiiIZdaGadaqaaiaaiAdacaGG % 7aGaaG4naiaacUdacaaI4aaacaGL7bGaayzFaaaaaa!44BD! x \in Z \Rightarrow x \in \left\{ {6;7;8} \right\}\)

Điều kiện:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCa % aaleqabaGaaG4maaaakiabgkHiTiaaiodacaWG4bGaeyOpa4JaaGim % aiabgsDiBlaadIhadaqadaqaaiaadIhadaahaaWcbeqaaiaaikdaaa % GccqGHsislcaaIZaaacaGLOaGaayzkaaGaeyOpa4JaaGimaiabgsDi % BlaadIhadaqadaqaaiaadIhacqGHsisldaGcaaqaaiaaiodaaSqaba % aakiaawIcacaGLPaaadaqadaqaaiaadIhacqGHRaWkdaGcaaqaaiaa % iodaaSqabaaakiaawIcacaGLPaaacqGH+aGpcaaIWaGaeyi1HS9aam % qaaqaabeqaaiabgkHiTmaakaaabaGaaG4maaWcbeaakiabgYda8iaa % dIhacqGH8aapcaaIWaaabaGaamiEaiabg6da+maakaaabaGaaG4maa % WcbeaaaaGccaGLBbaaaaa!6019! {x^3} - 3x > 0 \Leftrightarrow x\left( {{x^2} - 3} \right) > 0 \Leftrightarrow x\left( {x - \sqrt 3 } \right)\left( {x + \sqrt 3 } \right) > 0 \Leftrightarrow \left[ \begin{array}{l} - \sqrt 3 < x < 0\\ x > \sqrt 3 \end{array} \right.\)

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaWGLbWaaeWaaeaacaaIZaGaamiEamaaCaaaleqabaGa % aGOmaaaakiabgkHiTiaaiodaaiaawIcacaGLPaaadaqadaqaaiaadI % hadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaGaamiEaaGaayjk % aiaawMcaamaaCaaaleqabaGaamyzaiabgkHiTiaaigdaaaaaaa!484D! y' = e\left( {3{x^2} - 3} \right){\left( {{x^3} - 3x} \right)^{e - 1}}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yEaiaacEcacqGH9aqpcaaIWaGaeyi1HS9aaeWaaeaacaaIZaGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiodaaiaawIcacaGLPa % aadaqadaqaaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaI % ZaGaamiEaaGaayjkaiaawMcaamaaCaaaleqabaGaamyzaiabgkHiTi % aaigdaaaGccqGH9aqpcaaIWaGaeyi1HSTaaG4maiaadIhadaahaaWc % beqaaiaaikdaaaGccqGHsislcaaIZaGaeyypa0JaaGimaiabgsDiBp % aadeaaeaqabeaacaWG4bGaeyypa0JaeyOeI0IaaGymaaqaaiaadIha % cqGH9aqpcaaIXaaaaiaawUfaaaaa!60DA! \Rightarrow y' = 0 \Leftrightarrow \left( {3{x^2} - 3} \right){\left( {{x^3} - 3x} \right)^{e - 1}} = 0 \Leftrightarrow 3{x^2} - 3 = 0 \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = 1 \end{array} \right.\)

Ta có bảng xét dấu:

Dựa vào bảng xét dấu của hàm số ta thấy đạo hàm của hàm số chỉ đổi dấu qua 1 điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgkHiTiaaigdacqGHshI3aaa!3BFC! x = - 1 \Rightarrow \) hàm số có 1 điểm cực trị.

Ta có công thức tính thể tích hình phẳng đã cho là:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9iabec8aWnaapehabaWaaeWaaeaacaaIYaWaaWbaaSqabeaacaWG % 4baaaaGccaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaamizai % aadIhacqGH9aqpcqaHapaCdaWdXbqaaiaaisdadaahaaWcbeqaaiaa % dIhaaaGccaWGKbGaamiEaaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIi % paaSqaaiaaicdaaeaacaaIYaaaniabgUIiYdaaaa!4E3D! V = \pi \int\limits_0^2 {{{\left( {{2^x}} \right)}^2}dx = \pi \int\limits_0^2 {{4^x}dx} } \)

Dựa vào đồ thị hàm số ta có hàm số  y = f(x) đồng biến trên các khoảng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % GHsislcqGHEisPcaGG7aGaaGimaaGaayjkaiaawMcaaaaa!3B54! \left( { - \infty ;0} \right)\) và  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIYaGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawMcaaaaa!3B4B! \left( {2; + \infty } \right)\)

Hàm số y = f(x) nghịch biến trên (0;2)

Xét hàm số: y = -2f(x) ta có:  y' = -2f'(x)

Hàm số đồng biến  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaey % OeI0IaaGOmaiaadAgacaGGNaWaaeWaaeaacaWG4baacaGLOaGaayzk % aaGaeyyzImRaaGimaiabgsDiBlaadAgacaGGNaWaaeWaaeaacaWG4b % aacaGLOaGaayzkaaGaeyizImQaaGimaiabgsDiBlaaicdacqGHKjYO % caWG4bGaeyizImQaaGOmaaaa!51B5! \Leftrightarrow - 2f'\left( x \right) \ge 0 \Leftrightarrow f'\left( x \right) \le 0 \Leftrightarrow 0 \le x \le 2\)

Vậy hàm số y = -2f(x) đồng biến \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % iEaiabgIGiopaadmaabaGaaGimaiaacUdacaaIYaaacaGLBbGaayzx % aaaaaa!3EF8! \Leftrightarrow x \in \left[ {0;2} \right]\)

Điều kiện:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgc % Mi5kaaigdaaaa!3973! x \ne 1\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % iEaiabg2da9iaaigdaaaa!3B0F! \Rightarrow x = 1\) là đường TCĐ của đồ thị hàm số.

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHRaWkcqGHEisP % aeqaaOWaaSaaaeaacaWG4bGaey4kaSYaaOaaaeaacaWG4bWaaWbaaS % qabeaacaaIYaaaaOGaey4kaSIaaGymaaWcbeaaaOqaaiaadIhacqGH % sislcaaIXaaaaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaale % aacaWG4bGaeyOKH4Qaey4kaSIaeyOhIukabeaakmaalaaabaGaaGym % aiabgUcaRmaakaaabaGaaGymaiabgUcaRmaalaaabaGaaGymaaqaai % aadIhadaahaaWcbeqaaiaaikdaaaaaaaqabaaakeaacaaIXaGaeyOe % I0YaaSaaaeaacaaIXaaabaGaamiEaaaaaaGaeyypa0JaaGOmaaaa!5B27! \mathop {\lim }\limits_{x \to + \infty } \frac{{x + \sqrt {{x^2} + 1} }}{{x - 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 + \sqrt {1 + \frac{1}{{{x^2}}}} }}{{1 - \frac{1}{x}}} = 2\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yEaiabg2da9iaaikdaaaa!3B11! \Rightarrow y = 2\) là 1 đường TCN của đồ thị hàm số.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHsislcqGHEisP % aeqaaOWaaSaaaeaacaWG4bGaey4kaSYaaOaaaeaacaWG4bWaaWbaaS % qabeaacaaIYaaaaOGaey4kaSIaaGymaaWcbeaaaOqaaiaadIhacqGH % sislcaaIXaaaaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaale % aacaWG4bGaeyOKH4QaeyOeI0IaeyOhIukabeaakmaalaaabaGaaGym % aiabgkHiTmaakaaabaGaaGymaiabgUcaRmaalaaabaGaaGymaaqaai % aadIhadaahaaWcbeqaaiaaikdaaaaaaaqabaaakeaacaaIXaGaeyOe % I0YaaSaaaeaacaaIXaaabaGaamiEaaaaaaGaeyypa0JaaGimaaaa!5B46! \mathop {\lim }\limits_{x \to - \infty } \frac{{x + \sqrt {{x^2} + 1} }}{{x - 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{1 - \sqrt {1 + \frac{1}{{{x^2}}}} }}{{1 - \frac{1}{x}}} = 0\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yEaiabg2da9iaaicdaaaa!3B0F! \Rightarrow y = 0\) là 1 đường TCN của đồ thị hàm số.

Vậy đồ thị hàm số đã cho có 2 TCN và 1 TCĐ

Dựa vào đồ thị hàm số ta thấy \(x_1\) là nghiệm của phương trình hoành độ giao điểm  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaadkgaaeqaaOGaamiEamaaBaaaleaacaaI % Xaaabeaakiabg2da9iaaiodacqGHuhY2caWG4bWaaSbaaSqaaiaaig % daaeqaaOGaeyypa0JaamOyamaaCaaaleqabaGaaG4maaaaaaa!44B3! {\log _b}{x_1} = 3 \Leftrightarrow {x_1} = {b^3}\)

Và \(x_2\) là nghiệm của phương trình hoành độ giao điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaadggaaeqaaOGaamiEamaaBaaaleaacaaI % Yaaabeaakiabg2da9iaaiodacqGHuhY2caWG4bWaaSbaaSqaaiaaik % daaeqaaOGaeyypa0JaamyyamaaCaaaleqabaGaaG4maaaaaaa!44B3! {\log _a}{x_2} = 3 \Leftrightarrow {x_2} = {a^3}\)

Theo đề bài ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIYaaabeaakiabg2da9iaaikdacaWG4bWaaSbaaSqaaiaa % igdaaeqaaOGaeyO0H4TaamyyamaaCaaaleqabaGaaG4maaaakiabg2 % da9iaaikdacaWGIbWaaWbaaSqabeaacaaIZaaaaOGaeyi1HS9aaSaa % aeaacaWGHbWaaWbaaSqabeaacaaIZaaaaaGcbaGaamOyamaaCaaale % qabaGaaG4maaaaaaGccqGH9aqpcaaIYaGaeyi1HS9aaSaaaeaacaWG % HbaabaGaamOyaaaacqGH9aqpdaGcbaqaaiaaikdaaSqaaiaaiodaaa % aaaa!521D! {x_2} = 2{x_1} \Rightarrow {a^3} = 2{b^3} \Leftrightarrow \frac{{{a^3}}}{{{b^3}}} = 2 \Leftrightarrow \frac{a}{b} = \sqrt[3]{2}\)

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaado % eacqGH9aqpdaGcaaqaaiaadgeacaWGcbWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaamOqaiaadoeadaahaaWcbeqaaiaaikdaaaaabeaaki % abg2da9maakaaabaGaamyyamaaCaaaleqabaGaaGOmaaaakiabgUca % RiaaisdacaWGHbWaaWbaaSqabeaacaaIYaaaaaqabaGccqGH9aqpca % WGHbWaaOaaaeaacaaI1aaaleqaaaaa!47AA! AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{a^2} + 4{a^2}} = a\sqrt 5 \) (định lý Pitago)

Xét tam giác ACC’ vuông tại C ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGdb % Gaam4qaiaacEcacqGH9aqpdaGcaaqaaiaadgeacaWGdbGaai4jamaa % CaaaleqabaGaaGOmaaaakiabgkHiTiaadgeacaWGdbWaaWbaaSqabe % aacaaIYaaaaaqabaGccqGH9aqpdaGcaaqaaiaaiAdacaWGHbWaaWba % aSqabeaacaaIYaaaaOGaeyOeI0IaaGynaiaadggadaahaaWcbeqaai % aaikdaaaaabeaakiabg2da9iaadggaaeaacqGHshI3caWGwbWaaSba % aSqaaiaadgeacaWGcbGaam4qaiaadseacaGGUaGaamyqaiaacEcaca % WGcbGaai4jaiaadoeacaGGNaGaamiraiaacEcaaeqaaOGaeyypa0Ja % am4qaiaadoeacaGGNaGaaiOlaiaadgeacaWGcbGaaiOlaiaadgeaca % WGebGaeyypa0Jaamyyaiaac6cacaWGHbGaaiOlaiaaikdacaWGHbGa % eyypa0JaaGOmaiaadggadaahaaWcbeqaaiaaiodaaaaaaaa!6739! \begin{array}{l} CC' = \sqrt {AC{'^2} - A{C^2}} = \sqrt {6{a^2} - 5{a^2}} = a\\ \Rightarrow {V_{ABCD.A'B'C'D'}} = CC'.AB.AD = a.a.2a = 2{a^3} \end{array}\)

Ta có:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcaaIWaaaaa!3BD0! f'\left( x \right) = 0\)

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuh % Y2daqadaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG % 4baacaGLOaGaayzkaaWaaeWaaeaacaWG4bGaeyOeI0IaaGOmaaGaay % jkaiaawMcaamaaCaaaleqabaGaaGOmaaaakmaabmaabaGaaGOmamaa % CaaaleqabaGaamiEaaaakiabgkHiTiaaisdaaiaawIcacaGLPaaacq % GH9aqpcaaIWaaabaGaeyi1HSTaamiEamaabmaabaGaamiEaiabgUca % RiaaigdaaiaawIcacaGLPaaadaqadaqaaiaadIhacqGHsislcaaIYa % aacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOWaaeWaaeaacaaI % YaWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaaGOmamaaCaaaleqaba % GaaGOmaaaaaOGaayjkaiaawMcaaiabg2da9iaaicdaaeaacqGHuhY2 % daWabaabaeqabaGaamiEaiabg2da9iaaicdaaeaacaWG4bGaey4kaS % IaaGymaiabg2da9iaaicdaaeaacaWG4bGaeyOeI0IaaGOmaiabg2da % 9iaaicdaaeaacaaIYaWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaaG % OmamaaCaaaleqabaGaaGOmaaaakiabg2da9iaaicdaaaGaay5waaGa % eyi1HS9aamqaaqaabeqaaiaadIhacqGH9aqpcaaIWaaabaGaamiEai % abg2da9iabgkHiTiaaigdaaeaacaWG4bGaeyypa0JaaGOmaaqaaiaa % dIhacqGH9aqpcaaIYaaaaiaawUfaaiabgsDiBpaadeaaeaqabeaaca % WG4bGaeyypa0JaaGimaiaaykW7caaMc8UaaiikaiaackgacaGGVbGa % aiyAaiaaykW7caaMc8UaaGymaiaacMcaaeaacaWG4bGaeyypa0Jaey % OeI0IaaGymaiaaykW7caaMc8UaaiikaiaackgacaGGVbGaaiyAaiaa % ykW7caaMc8UaaGymaiaacMcaaeaacaWG4bGaeyypa0JaaGOmaiaayk % W7caaMc8UaaiikaiaackgacaGGVbGaaiyAaiaaykW7caaMc8UaaG4m % aiaacMcaaaGaay5waaaaaaa!B032! \begin{array}{l} \Leftrightarrow \left( {{x^2} + x} \right){\left( {x - 2} \right)^2}\left( {{2^x} - 4} \right) = 0\\ \Leftrightarrow x\left( {x + 1} \right){\left( {x - 2} \right)^2}\left( {{2^x} - {2^2}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x + 1 = 0\\ x - 2 = 0\\ {2^x} - {2^2} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = - 1\\ x = 2\\ x = 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\,\,(bội\,\,1)\\ x = - 1\,\,(bội\,\,1)\\ x = 2\,\,(bội\,\,3) \end{array} \right. \end{array}\)

Ta thấy phương trình f'(x) = 0 có 3 nghiệm phân biệt và các nghiệm này đều là nghiệm bội lẻ nên hàm số y = f(x) có 3 điểm cực trị.

Ta có hình hộp ABCD.A'B'C'D' có các cạnh bằng a \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yqaiaaykW7caWGbbGaai4jaiabg2da9iaadggaaaa!3DFF! \Rightarrow A\,A' = a\) là đường sinh của hình trụ.

Bán kính đáy của hình trụ là  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9maalaaabaGaamyqaiaadoeaaeaacaaIYaaaaiabg2da9maalaaa % baGaamyyamaakaaabaGaaGOmaaWcbeaaaOqaaiaaikdaaaaaaa!3DC3! R = \frac{{AC}}{2} = \frac{{a\sqrt 2 }}{2}\)

Diện tích xung quanh của hình trụ là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWG4bGaamyCaaqabaGccqGH9aqpcaaIYaGaeqiWdaNaamOu % aiaadYgacqGH9aqpcaaIYaGaeqiWdaNaaiOlamaalaaabaGaamyyam % aakaaabaGaaGOmaaWcbeaaaOqaaiaaikdaaaGaaiOlaiaadggacqGH % 9aqpdaGcaaqaaiaaikdaaSqabaGccqaHapaCcaWGHbWaaWbaaSqabe % aacaaIYaaaaaaa!4C0B! {S_{xq}} = 2\pi Rl = 2\pi .\frac{{a\sqrt 2 }}{2}.a = \sqrt 2 \pi {a^2}\)

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWG6bGaey4kaSIaaG4m % aiabg2da9iaaicdacqGHuhY2daWabaabaeqabaGaamOEamaaBaaale % aacaaIXaaabeaakiabg2da9iaaigdacqGHRaWkdaGcaaqaaiaaikda % aSqabaGccaWGPbGaeyO0H49aaqWaaeaacaWG6bWaaSbaaSqaaiaaig % daaeqaaaGccaGLhWUaayjcSdGaeyypa0ZaaOaaaeaacaaIXaGaey4k % aSIaaGOmaaWcbeaakiabg2da9maakaaabaGaaG4maaWcbeaaaOqaai % aadQhadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaaIXaGaeyOeI0Ya % aOaaaeaacaaIYaaaleqaaOGaamyAaiabgkDiEpaaemaabaGaamOEam % aaBaaaleaacaaIYaaabeaaaOGaay5bSlaawIa7aiabg2da9maakaaa % baGaaGymaiabgUcaRiaaikdaaSqabaGccqGH9aqpdaGcaaqaaiaaio % daaSqabaaaaOGaay5waaaaaa!67D2! {z^2} - 2z + 3 = 0 \Leftrightarrow \left[ \begin{array}{l} {z_1} = 1 + \sqrt 2 i \Rightarrow \left| {{z_1}} \right| = \sqrt {1 + 2} = \sqrt 3 \\ {z_2} = 1 - \sqrt 2 i \Rightarrow \left| {{z_2}} \right| = \sqrt {1 + 2} = \sqrt 3 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aaq % WaaeaacaWG6bWaa0baaSqaaiaaigdaaeaacaaIZaaaaOGaaiOlaiaa % dQhadaqhaaWcbaGaaGOmaaqaaiaaisdaaaaakiaawEa7caGLiWoacq % GH9aqpdaabdaqaaiaadQhadaWgaaWcbaGaaGymaaqabaaakiaawEa7 % caGLiWoadaahaaWcbeqaaiaaiodaaaGccaGGUaWaaqWaaeaacaWG6b % WaaSbaaSqaaiaaikdaaeqaaaGccaGLhWUaayjcSdWaaWbaaSqabeaa % caaI0aaaaOGaeyypa0ZaaeWaaeaadaGcaaqaaiaaiodaaSqabaaaki % aawIcacaGLPaaadaahaaWcbeqaaiaaiodaaaGccaGGUaWaaeWaaeaa % daGcaaqaaiaaiodaaSqabaaakiaawIcacaGLPaaadaahaaWcbeqaai % aaisdaaaGccqGH9aqpdaqadaqaamaakaaabaGaaG4maaWcbeaaaOGa % ayjkaiaawMcaamaaCaaaleqabaGaaG4naaaakiabg2da9iaaikdaca % aI3aWaaOaaaeaacaaIZaaaleqaaaaa!5F83! \Rightarrow \left| {z_1^3.z_2^4} \right| = {\left| {{z_1}} \right|^3}.{\left| {{z_2}} \right|^4} = {\left( {\sqrt 3 } \right)^3}.{\left( {\sqrt 3 } \right)^4} = {\left( {\sqrt 3 } \right)^7} = 27\sqrt 3 \)

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaWG4bGaey4k % aSIaci4yaiaac+gacaGGZbWaaSaaaeaacqaHapaCcaWG4baabaGaaG % OmaaaacqGHshI3caWGMbGaai4jamaabmaabaGaamiEaaGaayjkaiaa % wMcaaiabg2da9iaaikdacqGHsisldaWcaaqaaiabec8aWbqaaiaaik % daaaGaci4CaiaacMgacaGGUbWaaSaaaeaacqaHapaCcaWG4baabaGa % aGOmaaaaaaa!556E! f\left( x \right) = 2x + \cos \frac{{\pi x}}{2} \Rightarrow f'\left( x \right) = 2 - \frac{\pi }{2}\sin \frac{{\pi x}}{2}\)

Vì  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % ymaiabgsMiJkGacohacaGGPbGaaiOBamaalaaabaGaeqiWdaNaamiE % aaqaaiaaikdaaaGaeyizImQaaGymaiabgsDiBlabgkHiTmaalaaaba % GaeqiWdahabaGaaGOmaaaacqGHKjYOdaWcaaqaaiabec8aWbqaaiaa % ikdaaaGaci4CaiaacMgacaGGUbWaaSaaaeaacqaHapaCcaWG4baaba % GaaGOmaaaacqGHKjYOdaWcaaqaaiabec8aWbqaaiaaikdaaaGaeyO0 % H4TaaGimaiabgYda8iaaikdacqGHsisldaWcaaqaaiabec8aWbqaai % aaikdaaaGaeyizImQaaGOmaiabgkHiTmaalaaabaGaeqiWdahabaGa % aGOmaaaaciGGZbGaaiyAaiaac6gadaWcaaqaaiabec8aWjaadIhaae % aacaaIYaaaaiabgsMiJkaaikdacqGHsisldaWcaaqaaiabec8aWbqa % aiaaikdaaaaaaa!7144! - 1 \le \sin \frac{{\pi x}}{2} \le 1 \Leftrightarrow - \frac{\pi }{2} \le \frac{\pi }{2}\sin \frac{{\pi x}}{2} \le \frac{\pi }{2} \Rightarrow 0 < 2 - \frac{\pi }{2} \le 2 - \frac{\pi }{2}\sin \frac{{\pi x}}{2} \le 2 - \frac{\pi }{2}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH+aGpcaaI % WaGaeyiaIiIaamiEaiabgIGiopaadmaabaGaeyOeI0IaaGOmaiaacU % dacaaIYaaacaGLBbGaayzxaaGaeyO0H4naaa!48F3! \Rightarrow f'\left( x \right) > 0\forall x \in \left[ { - 2;2} \right] \Rightarrow \) hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaWG4bGaey4k % aSIaci4yaiaac+gacaGGZbWaaSaaaeaacqaHapaCcaWG4baabaGaaG % Omaaaaaaa!435F! f\left( x \right) = 2x + \cos \frac{{\pi x}}{2}\) là hàm đồng biến trên [-2;2]

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3caWGMbWaaeWaaeaacqGHsislcaaIYaaacaGLOaGaayzkaaGaeyiz % ImQaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiabgsMiJkaadA % gadaqadaqaaiaaikdaaiaawIcacaGLPaaacqGHaiIicaWG4bGaeyic % I48aamWaaeaacqGHsislcaaIYaGaai4oaiaaikdaaiaawUfacaGLDb % aaaeaacqGHshI3daGabaabaeqabaGaamytaiabg2da9maaxababaGa % ciyBaiaacggacaGG4baaleaadaWadaqaaiabgkHiTiaaikdacaGG7a % GaaGOmaaGaay5waiaaw2faaaqabaGccaWGMbWaaeWaaeaacaWG4baa % caGLOaGaayzkaaGaeyypa0JaamOzamaabmaabaGaaGOmaaGaayjkai % aawMcaaiabg2da9iaaiodaaeaacaWGTbGaeyypa0ZaaCbeaeaaciGG % TbGaaiyAaiaac6gaaSqaamaadmaabaGaeyOeI0IaaGOmaiaacUdaca % aIYaaacaGLBbGaayzxaaaabeaakiaadAgadaqadaqaaiaadIhaaiaa % wIcacaGLPaaacqGH9aqpcaWGMbWaaeWaaeaacqGHsislcaaIYaaaca % GLOaGaayzkaaGaeyypa0JaeyOeI0IaaGynaaaacaGL7baaaeaacqGH % shI3caWGnbGaey4kaSIaamyBaiabg2da9iaaiodacqGHRaWkcaGGOa % GaeyOeI0IaaGynaiaacMcacqGH9aqpcqGHsislcaaIYaaaaaa!889A! \begin{array}{l} \Rightarrow f\left( { - 2} \right) \le f\left( x \right) \le f\left( 2 \right)\forall x \in \left[ { - 2;2} \right]\\ \Rightarrow \left\{ \begin{array}{l} M = \mathop {\max }\limits_{\left[ { - 2;2} \right]} f\left( x \right) = f\left( 2 \right) = 3\\ m = \mathop {\min }\limits_{\left[ { - 2;2} \right]} f\left( x \right) = f\left( { - 2} \right) = - 5 \end{array} \right.\\ \Rightarrow M + m = 3 + ( - 5) = - 2 \end{array}\)

Gọi O là giao điểm của AC và BD

SABCD là hình chóp đều  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % 4uaiaad+eacqGHLkIxdaqadaqaaiaadgeacaWGcbGaam4qaiaadsea % aiaawIcacaGLPaaaaaa!4055! \Rightarrow SO \bot \left( {ABCD} \right)\)

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGtbGaamyqaiaadkeaaiaawIcacaGLPaaacqGHPiYXdaqadaqaaiaa % dgeacaWGcbGaam4qaiaadseaaiaawIcacaGLPaaacqGH9aqpdaGada % qaaiaadgeacaWGcbaacaGL7bGaayzFaaaaaa!44EB! \left( {SAB} \right) \cap \left( {ABCD} \right) = \left\{ {AB} \right\}\)

Gọi M là trung điểm của AB.

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaad2 % eacqGHLkIxcaWGbbGaamOqamaabmaabaGaam4taiaad2eacaGGVaGa % ai4laiaadgeacaWGebGaaiilaiaadgeacaWGebGaeyyPI4Laamyqai % aadkeaaiaawIcacaGLPaaaaaa!4679! OM \bot AB\left( {OM//AD,AD \bot AB} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaad2 % eacqGHLkIxcaWGbbGaamOqaaaa!3ADC! SM \bot AB\) do \(\Delta SAB\) là tam giác cân tại S.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taey % iiIa9aaeWaaeaadaqadaqaaiaadofacaWGbbGaamOqaaGaayjkaiaa % wMcaaiaacYcadaqadaqaaiaadgeacaWGcbGaam4qaiaadseaaiaawI % cacaGLPaaaaiaawIcacaGLPaaacqGH9aqpcqGHGic0daqadaqaaiaa % dofacaWGnbGaaiilaiaad+eacaWGnbaacaGLOaGaayzkaaGaeyypa0 % JaeyiiIaTaam4uaiaad2eacaWGpbaaaa!520C! \Rightarrow \angle \left( {\left( {SAB} \right),\left( {ABCD} \right)} \right) = \angle \left( {SM,OM} \right) = \angle SMO\)

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaad2 % eacqGH9aqpdaGcaaqaaiaadofacaWGbbWaaWbaaSqabeaacaaIYaaa % aOGaeyOeI0IaamytaiaadgeadaahaaWcbeqaaiaaikdaaaaabeaaki % abg2da9maakaaabaGaaGynaiaadggadaahaaWcbeqaaiaaikdaaaGc % cqGHsislcaWGHbWaaWbaaSqabeaacaaIYaaaaaqabaGccqGH9aqpca % aIYaGaamyyaaaa!47D9! SM = \sqrt {S{A^2} - M{A^2}} = \sqrt {5{a^2} - {a^2}} = 2a\) (Định lý Pitago)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaad2 % eacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadgeacaWGebGa % eyypa0Jaamyyaaaa!3DA2! OM = \frac{1}{2}AD = a\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3ciGGJbGaai4BaiaacohacaWGtbGaamytaiaad+eacqGH9aqpdaWc % aaqaaiaad+eacaWGnbaabaGaam4uaiaad2eaaaGaeyypa0ZaaSaaae % aacaWGHbaabaGaaGOmaiaadggaaaGaeyypa0ZaaSaaaeaacaaIXaaa % baGaaGOmaaaaaeaacqGHshI3cqGHGic0caWGtbGaamytaiaad+eacq % GH9aqpcaaI2aGaaGimamaaCaaaleqabaGaaGimaaaaaaaa!521A! \begin{array}{l} \Rightarrow \cos SMO = \frac{{OM}}{{SM}} = \frac{a}{{2a}} = \frac{1}{2}\\ \Rightarrow \angle SMO = {60^0} \end{array}\)

Số các số tự nhiên 2 chữ số phân biệt là 9.9 = 81  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OBamaabmaabaGaeuyQdCfacaGLOaGaayzkaaGaeyypa0JaaGioaiaa % igdadaahaaWcbeqaaiaaikdaaaaaaa!3FC7! \Rightarrow n\left( \Omega \right) = {81^2}\)

Gọi A là biến cố: “Hai số được viết ra có ít nhất một chữ số chung”

TH1: Hai bạn cùng viết ra số giống nhau ; Có 81 cách

TH2: Bạn Công viết số có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaamOyaaaaaaa!37D2! \overline {ab} \) và bạn Thành viết số có dạng  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGIbGaamyyaaaaaaa!37D2! \overline {ba} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yyaiabgcMi5kaadkgacqGHGjsUcaaIWaGaeyO0H4naaa!40C3! \Rightarrow a \ne b \ne 0 \Rightarrow \) Có 9.8 = 72 cách.

TH3: Hai bạn chọn số chỉ có 1 chữ số trùng nhau.

+) Trùng số 0: Số cần viết có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaaGimaaaaaaa!37A5! \overline {a0} \), Công có 9 cách viết, Thành có 8 cách viết (Khác số Công viết)

 Có 9.8 = 72 cách.

+) Trùng số 1: Số cần viết có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaaGymaaaadaqadaqaaiaadggacqGHGjsUcaaIWaGaaiilaiaa % dggacqGHGjsUcaaIXaaacaGLOaGaayzkaaaaaa!40AE! \overline {a1} \left( {a \ne 0,a \ne 1} \right)\), hoặc  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % aIXaGaamOyaaaadaqadaqaaiaadkgacqGHGjsUcaaIXaaacaGLOaGa % ayzkaaaaaa!3C99! \overline {1b} \left( {b \ne 1} \right)\)

Nếu Công viết số 10, khi đó Thành có 8 cách viết số có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaaGymaaaadaqadaqaaiaadggacqGHGjsUcaaIWaGaaiilaiaa % dggacqGHGjsUcaaIXaaacaGLOaGaayzkaaaaaa!40AE! \overline {a1} \left( {a \ne 0,a \ne 1} \right)\) và 8 cách viết số có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % aIXaGaamOyaaaadaqadaqaaiaadkgacqGHGjsUcaaIXaaacaGLOaGa % ayzkaaaaaa!3C99! \overline {1b} \left( {b \ne 1} \right)\) . Có 16 cách.

Nếu Công viết số có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % aIXaGaamOyaaaadaqadaqaaiaadkgacqGHGjsUcaaIWaGaaiilaiaa % dkgacqGHGjsUcaaIXaaacaGLOaGaayzkaaGaeyO0H4naaa!430E! \overline {1b} \left( {b \ne 0,b \ne 1} \right) \Rightarrow \) Công có 8 cách viết, khi đó Thành có 7 cách viết số có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaaGymaaaadaqadaqaaiaadggacqGHGjsUcaaIWaGaaiilaiaa % dggacqGHGjsUcaaIXaaacaGLOaGaayzkaaaaaa!40AE! \overline {a1} \left( {a \ne 0,a \ne 1} \right)\)

 và 8 cách viết số có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % aIXaGaamOyaaaadaqadaqaaiaadkgacqGHGjsUcaaIXaaacaGLOaGa % ayzkaaaaaa!3C99! \overline {1b} \left( {b \ne 1} \right)\)

Có 8 (7 + 8) = 120 cách.

Nếu Công viết số có dạng  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaaGymaaaadaqadaqaaiaadggacqGHGjsUcaaIWaGaaiilaiaa % dggacqGHGjsUcaaIXaaacaGLOaGaayzkaaaaaa!40AE! \overline {a1} \left( {a \ne 0,a \ne 1} \right)\)  Công có 8 cách viết, khi đó Thành có 7 cách viết số có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbGaaGymaaaadaqadaqaaiaadggacqGHGjsUcaaIWaGaaiilaiaa % dggacqGHGjsUcaaIXaaacaGLOaGaayzkaaaaaa!40AE! \overline {a1} \left( {a \ne 0,a \ne 1} \right)\) và 8 cách viết số có dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % aIXaGaamOyaaaadaqadaqaaiaadkgacqGHGjsUcaaIXaaacaGLOaGa % ayzkaaaaaa!3C99! \overline {1b} \left( {b \ne 1} \right)\)

Có 8 (7 + 8) = 120 cách.

Có 256 cách viết trùng số 1.

Tương tự cho các trường hợp trùng số 2,3,4,5,6,7,8,9

 \(\Rightarrow n\left( A \right) = 81 + 72 + 72 + 256.9 = 2529\)

Vậy  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaabm % aabaGaamyqaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGOmaiaa % iwdacaaIYaGaaGyoaaqaaiaaiIdacaaIXaWaaWbaaSqabeaacaaIYa % aaaaaakiabg2da9maalaaabaGaaGOmaiaaiIdacaaIXaaabaGaaG4n % aiaaikdacaaI5aaaaaaa!4527! P\left( A \right) = \frac{{2529}}{{{{81}^2}}} = \frac{{281}}{{729}}\)

Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiaadw % gadaahaaWcbeqaaiaadIhaaaaaaa!3905! x{e^x}\) là một nguyên hàm của hàm số f(-x) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaamyzamaaCaaaleqabaGaamiEaaaaaOGaayjkaiaawMcaaiaa % cEcacqGH9aqpcaWGMbWaaeWaaeaacqGHsislcaWG4baacaGLOaGaay % zkaaaaaa!40A7! \left( {x{e^x}} \right)' = f\left( { - x} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % OzamaabmaabaGaeyOeI0IaamiEaaGaayjkaiaawMcaaiabg2da9iaa % dwgadaahaaWcbeqaaiaadIhaaaGccqGHRaWkcaWG4bGaamyzamaaCa % aaleqabaGaamiEaaaakiabg2da9iaadwgadaahaaWcbeqaaiaadIha % aaGcdaqadaqaaiaaigdacqGHRaWkcaWG4baacaGLOaGaayzkaaaaaa!4B16! \Leftrightarrow f\left( { - x} \right) = {e^x} + x{e^x} = {e^x}\left( {1 + x} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OzamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadwgadaah % aaWcbeqaaiabgkHiTiaadIhaaaGcdaqadaqaaiaaigdacqGHsislca % WG4baacaGLOaGaayzkaaaaaa!4401! \Rightarrow f\left( x \right) = {e^{ - x}}\left( {1 - x} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcqGH % sislcaWGLbWaaWbaaSqabeaacqGHsislcaWG4baaaOWaaeWaaeaaca % aIXaGaeyOeI0IaamiEaaGaayjkaiaawMcaaiabgkHiTiaadwgadaah % aaWcbeqaaiabgkHiTiaadIhaaaGccqGH9aqpcqGHsislcaWGLbWaaW % baaSqabeaacqGHsislcaWG4baaaOWaaeWaaeaacaaIYaGaeyOeI0Ia % amiEaaGaayjkaiaawMcaaiabg2da9maabmaabaGaamiEaiabgkHiTi % aaikdaaiaawIcacaGLPaaacaWGLbWaaWbaaSqabeaacqGHsislcaWG % 4baaaaaa!5AF4! \Rightarrow f'\left( x \right) = - {e^{ - x}}\left( {1 - x} \right) - {e^{ - x}} = - {e^{ - x}}\left( {2 - x} \right) = \left( {x - 2} \right){e^{ - x}}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3caWGMbGaai4jamaabmaabaGaamiEaaGaayjkaiaawMcaaiaadwga % daahaaWcbeqaaiaadIhaaaGccqGH9aqpdaqadaqaaiaadIhacqGHsi % slcaaIYaaacaGLOaGaayzkaaGaamyzamaaCaaaleqabaGaeyOeI0Ia % amiEaaaakiaac6cacaWGLbWaaWbaaSqabeaacaWG4baaaOGaeyypa0 % JaamiEaiabgkHiTiaaikdaaeaacqGHshI3caWGgbWaaeWaaeaacaWG % 4baacaGLOaGaayzkaaGaeyypa0Zaa8qaaeaacaWGMbWaaeWaaeaaca % WG4baacaGLOaGaayzkaaGaamizaiaadIhacqGH9aqpdaWdbaqaamaa % bmaabaGaamiEaiabgkHiTiaaikdaaiaawIcacaGLPaaacaWGKbGaam % iEaiabg2da9maalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaOqa % aiaaikdaaaGaeyOeI0IaaGOmaiaadIhacqGHRaWkcaWGdbaaleqabe % qdcqGHRiI8aaWcbeqab0Gaey4kIipaaOqaaiaadAeadaqadaqaaiaa % icdaaiaawIcacaGLPaaacqGH9aqpcaaIXaGaeyO0H4Taam4qaiabg2 % da9iaaigdacqGHshI3caWGgbWaaeWaaeaacaWG4baacaGLOaGaayzk % aaGaeyypa0ZaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGcba % GaaGOmaaaacqGHsislcaaIYaGaamiEaiabgUcaRiaaigdaaeaacqGH % shI3caWGgbWaaeWaaeaacqGHsislcaaIXaaacaGLOaGaayzkaaGaey % ypa0ZaaSaaaeaadaqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaa % daahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaiabgkHiTiaaikdada % qadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGHRaWkcaaIXaGa % eyypa0ZaaSaaaeaacaaI3aaabaGaaGOmaaaaaaaa!9949! \begin{array}{l} \Rightarrow f'\left( x \right){e^x} = \left( {x - 2} \right){e^{ - x}}.{e^x} = x - 2\\ \Rightarrow F\left( x \right) = \int {f\left( x \right)dx = \int {\left( {x - 2} \right)dx = \frac{{{x^2}}}{2} - 2x + C} } \\ F\left( 0 \right) = 1 \Rightarrow C = 1 \Rightarrow F\left( x \right) = \frac{{{x^2}}}{2} - 2x + 1\\ \Rightarrow F\left( { - 1} \right) = \frac{{{{\left( { - 1} \right)}^2}}}{2} - 2\left( { - 1} \right) + 1 = \frac{7}{2} \end{array}\)

Đặt hệ trục tọa độ như hình vẽ, chọn a = 1. Khi đó ta có:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaaGimaiaacUdacaaIWaGaai4oaiaaicdaaiaawIcacaGLPaaa % caGGSaGaamOqamaabmaabaGaaGOmaiaacUdacaaIWaGaai4oaiaaic % daaiaawIcacaGLPaaacaGGSaGaam4qamaabmaabaGaaGOmaiaacUda % caaIXaGaai4oaiaaicdaaiaawIcacaGLPaaacaGGSaGaamiramaabm % aabaGaaGimaiaacUdacaaIXaGaai4oaiaaicdaaiaawIcacaGLPaaa % caGGSaGaam4uamaabmaabaGaaGimaiaacUdacaaIWaGaai4oaiaaio % daaiaawIcacaGLPaaaaaa!56BC! A\left( {0;0;0} \right),B\left( {2;0;0} \right),C\left( {2;1;0} \right),D\left( {0;1;0} \right),S\left( {0;0;3} \right)\)

M là trung điểm cạnh  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qaiaads % eacqGHshI3caWGnbWaaeWaaeaacaaIXaGaai4oaiaaigdacaGG7aGa % aGimaaGaayjkaiaawMcaaaaa!3FEB! CD \Rightarrow M\left( {1;1;0} \right)\)

Ta có :\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGtbGaam4qaaGaay51GaGaeyypa0ZaaeWaaeaacqGHsislcaaIYaGa % ai4oaiabgkHiTiaaigdacaGG7aGaaG4maaGaayjkaiaawMcaaiaacU % dadaWhcaqaaiaadkeacaWGnbaacaGLxdcacqGH9aqpdaqadaqaaiab % gkHiTiaaigdacaGG7aGaaGymaiaacUdacaaIWaaacaGLOaGaayzkaa % Gaai4oamaaFiaabaGaam4uaiaadkeaaiaawEniaiabg2da9maabmaa % baGaaGOmaiaacUdacaaIWaGaai4oaiabgkHiTiaaiodaaiaawIcaca % GLPaaacqGHshI3daWadaqaamaaFiaabaGaam4uaiaadoeaaiaawEni % aiaacUdadaWhcaqaaiaadkeacaWGnbaacaGLxdcaaiaawUfacaGLDb % aacqGH9aqpdaqadaqaaiabgkHiTiaaiodacaGG7aGaeyOeI0IaaG4m % aiaacUdacqGHsislcaaIZaaacaGLOaGaayzkaaaaaa!6C92! \overrightarrow {SC} = \left( { - 2; - 1;3} \right);\overrightarrow {BM} = \left( { - 1;1;0} \right);\overrightarrow {SB} = \left( {2;0; - 3} \right) \Rightarrow \left[ {\overrightarrow {SC} ;\overrightarrow {BM} } \right] = \left( { - 3; - 3; - 3} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaam4uaiaadoeacaGG7aGaamOqaiaad2eaaiaawIcacaGLPaaa % cqGH9aqpdaWcaaqaamaaemaabaWaamWaaeaadaWhcaqaaiaadofaca % WGdbaacaGLxdcacaGG7aWaa8HaaeaacaWGcbGaamytaaGaay51Gaaa % caGLBbGaayzxaaGaaiOlamaaFiaabaGaam4uaiaadkeaaiaawEniaa % Gaay5bSlaawIa7aaqaamaaemaabaWaamWaaeaadaWhcaqaaiaadofa % caWGdbaacaGLxdcacaGG7aWaa8HaaeaacaWGcbGaamytaaGaay51Ga % aacaGLBbGaayzxaaaacaGLhWUaayjcSdaaaiabg2da9maalaaabaWa % aqWaaeaacqGHsislcaaIZaGaaiOlaiaaikdacqGHsislcaaIZaGaai % OlaiaaicdacqGHRaWkdaqadaqaaiabgkHiTiaaiodaaiaawIcacaGL % PaaacaGGUaWaaeWaaeaacqGHsislcaaIZaaacaGLOaGaayzkaaaaca % GLhWUaayjcSdaabaWaaOaaaeaadaqadaqaaiabgkHiTiaaiodaaiaa % wIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkdaqadaqaai % abgkHiTiaaiodaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGc % cqGHRaWkdaqadaqaaiabgkHiTiaaiodaaiaawIcacaGLPaaadaahaa % WcbeqaaiaaikdaaaaabeaaaaGccqGH9aqpdaWcaaqaaiaaiodaaeaa % caaIZaWaaOaaaeaacaaIZaaaleqaaaaakiabg2da9maalaaabaWaaO % aaaeaacaaIZaaaleqaaaGcbaGaaG4maaaaaaa!812E! \Rightarrow d\left( {SC;BM} \right) = \frac{{\left| {\left[ {\overrightarrow {SC} ;\overrightarrow {BM} } \right].\overrightarrow {SB} } \right|}}{{\left| {\left[ {\overrightarrow {SC} ;\overrightarrow {BM} } \right]} \right|}} = \frac{{\left| { - 3.2 - 3.0 + \left( { - 3} \right).\left( { - 3} \right)} \right|}}{{\sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( { - 3} \right)}^2}} }} = \frac{3}{{3\sqrt 3 }} = \frac{{\sqrt 3 }}{3}\)

Phương pháp:

Hàm số y = f(x) đồng biến trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGHbGaai4oaiaadkgaaiaawIcacaGLPaaacqGHuhY2caWGMbGaai4j % amaabmaabaGaamiEaaGaayjkaiaawMcaaiabgwMiZkaaicdacqGHai % IicaWG4bGaeyicI48aaeWaaeaacaWGHbGaai4oaiaadkgaaiaawIca % caGLPaaaaaa!4A67! \left( {a;b} \right) \Leftrightarrow f'\left( x \right) \ge 0\forall x \in \left( {a;b} \right)\) và bằng 0 tại hữu hạn điểm.

Cách giải:

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcqGHsislcaaIYaGaamOzaiaacEcadaqadaqaaiaaigda % cqGHsislcaaIYaGaamiEaaGaayjkaiaawMcaaaaa!40CC! y' = - 2f'\left( {1 - 2x} \right)\)

Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaaigdacqGHshI3caWG5bGaai4jamaabmaabaGaaGymaaGaayjk % aiaawMcaaiabg2da9iabgkHiTiaaikdacaWGMbGaai4jamaabmaaba % GaeyOeI0IaaGymaaGaayjkaiaawMcaaiabg6da+iaaicdacqGHshI3 % aaa!4A91! x = 1 \Rightarrow y'\left( 1 \right) = - 2f'\left( { - 1} \right) > 0 \Rightarrow \) Loại đáp án B, C, D.

Chọn A.

Chú ý: Ngoài phương pháp thử HS có thể lập BXD y’, tuy nhiên trong bài tập này, thử là phương pháp tối ưu nhất.

Theo bài ra ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG6b % Gaey4kaSIaaGOmaiabg2da9iaadMgacaWG3bGaeyO0H4Taam4Daiab % g2da9maalaaabaGaamOEaiabgUcaRiaaikdaaeaacaWGPbaaaaqaam % aaemaabaGaam4DaiabgkHiTiaadMgaaiaawEa7caGLiWoacqGH9aqp % caaIYaGaeyO0H49aaqWaaeaadaWcaaqaaiaadQhacqGHRaWkcaaIYa % aabaGaamyAaaaacqGHsislcaWGPbaacaGLhWUaayjcSdGaeyypa0Ja % aGOmaiabgsDiBpaaemaabaGaamOEaiabgUcaRiaaikdacqGHRaWkca % aIXaaacaGLhWUaayjcSdGaeyypa0JaaGOmaiabgsDiBpaaemaabaGa % amOEaiabgUcaRiaaiodaaiaawEa7caGLiWoacqGH9aqpcaaIYaaaaa % a!6D4C! \begin{array}{l} z + 2 = iw \Rightarrow w = \frac{{z + 2}}{i}\\ \left| {w - i} \right| = 2 \Rightarrow \left| {\frac{{z + 2}}{i} - i} \right| = 2 \Leftrightarrow \left| {z + 2 + 1} \right| = 2 \Leftrightarrow \left| {z + 3} \right| = 2 \end{array}\)

 Tập hợp các điểm biểu diễn số phức z là đường tròn I(-3;0) bán kính R = 2

Gọi M là điểm biểu diễn số phức z, dựa vào hình vẽ ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaamaaemaabaGaamOEaaGaay5bSlaawIa7amaaBaaaleaaciGGTbGa % aiyAaiaac6gaaeqaaOGaeyi1HSTaam4taiaad2eadaWgaaWcbaGaci % yBaiaacMgacaGGUbaabeaakiabgsDiBlaad2eadaqadaqaaiabgkHi % TiaaigdacaGG7aGaaGimaaGaayjkaiaawMcaaiabgkDiElaadQhada % WgaaWcbaGaaGymaaqabaGccqGH9aqpcqGHsislcaaIXaaabaWaaqWa % aeaacaWG6baacaGLhWUaayjcSdWaaSbaaSqaaiGac2gacaGGHbGaai % iEaaqabaGccqGHuhY2caWGpbGaamytamaaBaaaleaaciGGTbGaaiyy % aiaacIhaaeqaaOGaeyi1HSTaamytamaabmaabaGaeyOeI0IaaGynai % aacUdacaaIWaaacaGLOaGaayzkaaGaeyO0H4TaamOEamaaBaaaleaa % caaIYaaabeaakiabg2da9iabgkHiTiaaiwdaaaGaay5EaaGaeyO0H4 % 9aaqWaaeaacaWG6bWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamOE % amaaBaaaleaacaaIYaaabeaaaOGaay5bSlaawIa7aiabg2da9iaaiA % daaaa!7D36! \left\{ \begin{array}{l} {\left| z \right|_{\min }} \Leftrightarrow O{M_{\min }} \Leftrightarrow M\left( { - 1;0} \right) \Rightarrow {z_1} = - 1\\ {\left| z \right|_{\max }} \Leftrightarrow O{M_{\max }} \Leftrightarrow M\left( { - 5;0} \right) \Rightarrow {z_2} = - 5 \end{array} \right. \Rightarrow \left| {{z_1} + {z_2}} \right| = 6\)

Đáp án A: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTiaadAgadaqadaqaaiaadIhacqGHRaWkcaaIXaaacaGL % OaGaayzkaaGaeyOeI0IaaGymaiabg2da9iabgkHiTiaadIhadaahaa % WcbeqaaiaaiodaaaGccqGHsislcaaIZaWaaeWaaeaacaWG4bGaey4k % aSIaaGymaaGaayjkaiaawMcaaiabgUcaRiaaiodacqGHsislcaaIXa % Gaeyypa0JaeyOeI0IaamiEamaaCaaaleqabaGaaG4maaaakiabgkHi % TiaaiodacaWG4bGaeyOeI0IaaGymaaaa!54C6! y = - f\left( {x + 1} \right) - 1 = - {x^3} - 3\left( {x + 1} \right) + 3 - 1 = - {x^3} - 3x - 1\). Đồ thị hàm số đi qua điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaai4oaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGHshI3aaa!3CFB! \left( {0; - 1} \right) \Rightarrow \) Loại.

Đáp án B: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTiaadAgadaqadaqaaiaadIhacqGHRaWkcaaIXaaacaGL % OaGaayzkaaGaey4kaSIaaGymaiabg2da9iabgkHiTiaadIhadaahaa % WcbeqaaiaaiodaaaGccqGHsislcaaIZaWaaeWaaeaacaWG4bGaey4k % aSIaaGymaaGaayjkaiaawMcaaiabgUcaRiaaiodacqGHRaWkcaaIXa % Gaeyypa0JaeyOeI0IaamiEamaaCaaaleqabaGaaG4maaaakiabgUca % RiaaiodacaWG4bGaey4kaSIaaGymaaaa!549A! y = - f\left( {x + 1} \right) + 1 = - {x^3} - 3\left( {x + 1} \right) + 3 + 1 = - {x^3} + 3x + 1\). Đồ thị hàm số đi qua điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaai4oaiaaigdaaiaawIcacaGLPaaacqGHshI3aaa!3C0E! \left( {0;1} \right) \Rightarrow \) Đáp án B có thể đúng.

Đáp án C: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTmaabmaabaGaamiEaiabgkHiTiaaikdaaiaawIcacaGL % PaaadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaWaaeWaaeaaca % WG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaaiabgkHiTiaaigdacqGH % 9aqpcqGHsislcaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaaG % OnaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIXaGaaGyn % aiaadIhacqGHRaWkcaaIXaGaaGimaiabg2da9iaaicdaaaa!5479! y = - {\left( {x - 2} \right)^3} - 3\left( {x - 1} \right) - 1 = - {x^3} + 6{x^2} - 15x + 10 = 0\). Đồ thị hàm số đi qua điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaai4oaiaaigdacaaIWaaacaGLOaGaayzkaaGaeyO0H4naaa!3CC8! \left( {0;10} \right) \Rightarrow \) Loại.

Đáp án D:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTmaabmaabaGaamiEaiabgkHiTiaaikdaaiaawIcacaGL % PaaadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaWaaeWaaeaaca % WG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaaiabgUcaRiaaigdacqGH % 9aqpcqGHsislcaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaaG % OnaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIXaGaaGyn % aiaadIhacqGHRaWkcaaIXaGaaGOmaiabg2da9iaaicdaaaa!5470! y = - {\left( {x - 2} \right)^3} - 3\left( {x - 1} \right) + 1 = - {x^3} + 6{x^2} - 15x + 12 = 0\) . Đồ thị hàm số đi qua điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaai4oaiaaigdacaaIYaaacaGLOaGaayzkaaGaeyO0H4naaa!3CCA! \left( {0;12} \right) \Rightarrow \) Loại.

Dựa vào dữ kiện bài toán và hình vẽ suy ra  Hình trụ có chiều cao h = 2r và bán kính đáy  R = 2r

 Thể tích khối trụ là  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9iabec8aWnaabmaabaGaaGOmaiaadkhaaiaawIcacaGLPaaadaah % aaWcbeqaaiaaikdaaaGccaaIYaGaamOCaiabg2da9iaaiIdacqaHap % aCcaWGYbWaaWbaaSqabeaacaaIZaaaaOGaeyypa0JaaGymaiaaikda % caaIWaGaeyi1HSTaamOCamaaCaaaleqabaGaaG4maaaakiabg2da9m % aalaaabaGaaGymaiaaikdacaaIWaaabaGaaGioaiabec8aWbaacqGH % 9aqpdaWcaaqaaiaaigdacaaI1aaabaGaeqiWdahaaaaa!5675! V = \pi {\left( {2r} \right)^2}2r = 8\pi {r^3} = 120 \Leftrightarrow {r^3} = \frac{{120}}{{8\pi }} = \frac{{15}}{\pi }\)

Vậy thể tích mỗi khối cầu là  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGJbaabeaakiabg2da9maalaaabaGaaGinaaqaaiaaioda % aaGaeqiWdaNaamOCamaaCaaaleqabaGaaG4maaaakiabg2da9maala % aabaGaaGinaaqaaiaaiodaaaGaeqiWdaNaaiOlamaalaaabaGaaGym % aiaaiwdaaeaacqaHapaCaaGaeyypa0JaaGOmaiaaicdadaqadaqaai % aadogacaWGTbWaaWbaaSqabeaacaaIZaaaaaGccaGLOaGaayzkaaaa % aa!4D40! {V_c} = \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi .\frac{{15}}{\pi } = 20\left( {c{m^3}} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapehabaWaaSaaaeaaciGGJbGaai4BaiaacohadaahaaWcbeqa % aiaaikdaaaGccaWG4bGaey4kaSIaci4CaiaacMgacaGGUbGaaGPaVl % aadIhaciGGJbGaai4BaiaacohacaWG4bGaey4kaSIaaGymaaqaaiGa % cogacaGGVbGaai4CamaaCaaaleqabaGaaGinaaaakiaadIhacqGHRa % WkciGGZbGaaiyAaiaac6gacaaMc8UaamiEaiGacogacaGGVbGaai4C % amaaCaaaleqabaGaaG4maaaakiaadIhaaaGaamizaiaadIhaaSqaam % aalaaabaGaeqiWdahabaGaaGinaaaaaeaadaWcaaqaaiabec8aWbqa % aiaaiodaaaaaniabgUIiYdGccqGH9aqpdaWdXbqaamaalaaabaGaaG % ymaiabgUcaRiGacshacaGGHbGaaiOBaiaaykW7caWG4bGaey4kaSIa % aGymaiabgUcaRiGacshacaGGHbGaaiOBamaaCaaaleqabaGaaGOmaa % aakiaadIhaaeaaciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaikda % aaGccaWG4bWaaeWaaeaacaaIXaGaey4kaSIaciiDaiaacggacaGGUb % GaaGPaVlaadIhaaiaawIcacaGLPaaaaaGaamizaiaadIhacqGH9aqp % daWdXbqaamaalaaabaGaciiDaiaacggacaGGUbWaaWbaaSqabeaaca % aIYaaaaOGaamiEaiabgUcaRiGacshacaGGHbGaaiOBaiaaykW7caWG % 4bGaey4kaSIaaGOmaaqaaiGacogacaGGVbGaai4CamaaCaaaleqaba % GaaGOmaaaakiaadIhadaqadaqaaiaaigdacqGHRaWkciGG0bGaaiyy % aiaac6gacaaMc8UaamiEaaGaayjkaiaawMcaaaaacaWGKbGaamiEaa % WcbaWaaSaaaeaacqaHapaCaeaacaaI0aaaaaqaamaalaaabaGaeqiW % dahabaGaaG4maaaaa0Gaey4kIipaaSqaamaalaaabaGaeqiWdahaba % GaaGinaaaaaeaadaWcaaqaaiabec8aWbqaaiaaiodaaaaaniabgUIi % Ydaaaa!ABCB! I = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\cos }^2}x + \sin \,x\cos x + 1}}{{{{\cos }^4}x + \sin \,x{{\cos }^3}x}}dx} = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{1 + \tan \,x + 1 + {{\tan }^2}x}}{{{{\cos }^2}x\left( {1 + \tan \,x} \right)}}dx = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\tan }^2}x + \tan \,x + 2}}{{{{\cos }^2}x\left( {1 + \tan \,x} \right)}}dx} } \)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iGacshacaGGHbGaaiOBaiaaykW7caWG4baaaa!3D4C! t = \tan \,x\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % izaiaadshacqGH9aqpdaWcaaqaaiaaigdaaeaaciGGJbGaai4Baiaa % cohadaahaaWcbeqaaiaaikdaaaGccaWG4baaaiaadsgacaWG4baaaa!42AD! \Rightarrow dt = \frac{1}{{{{\cos }^2}x}}dx\) Đổi cận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaisdaaaGaeyO0 % H4TaamiDaiabg2da9iaaigdaaeaacaWG4bGaeyypa0ZaaSaaaeaacq % aHapaCaeaacaaIZaaaaiabgkDiElaadshacqGH9aqpdaGcaaqaaiaa % iodaaSqabaaaaOGaay5Eaaaaaa!4A85! \left\{ \begin{array}{l} x = \frac{\pi }{4} \Rightarrow t = 1\\ x = \frac{\pi }{3} \Rightarrow t = \sqrt 3 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ysaiabg2da9maapehabaWaaSaaaeaacaWG0bWaaWbaaSqabeaacaaI % YaaaaOGaey4kaSIaamiDaiabgUcaRiaaikdaaeaacaWG0bGaey4kaS % IaaGymaaaacaWGKbGaamiDaiabg2da9maapehabaWaaeWaaeaacaWG % 0bGaey4kaSYaaSaaaeaacaaIYaaabaGaamiDaiabgUcaRiaaigdaaa % aacaGLOaGaayzkaaGaamizaiaadshaaSqaaiaaigdaaeaadaGcaaqa % aiaaiodaaWqabaaaniabgUIiYdaaleaacaaIXaaabaWaaOaaaeaaca % aIZaaameqaaaqdcqGHRiI8aaaa!55BA! \Rightarrow I = \int\limits_1^{\sqrt 3 } {\frac{{{t^2} + t + 2}}{{t + 1}}dt = \int\limits_1^{\sqrt 3 } {\left( {t + \frac{2}{{t + 1}}} \right)dt} } \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpdaWcaaqaaiaadshadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaa % aiabgUcaRiaaikdaciGGSbGaaiOBamaaemaabaGaamiDaiabgUcaRi % aaigdaaiaawEa7caGLiWoadaabbaabaeqabaWaaWbaaSqabeaadaGc % aaqaaiaaiodaaWqabaaaaaGcbaWaaSbaaSqaaiaaigdaaeqaaaaaki % aawEa7aiabg2da9maalaaabaGaaG4maaqaaiaaikdaaaGaey4kaSIa % aGOmaiGacYgacaGGUbWaaeWaaeaadaGcaaqaaiaaiodaaSqabaGccq % GHRaWkcaaIXaaacaGLOaGaayzkaaGaeyOeI0YaaSaaaeaacaaIXaaa % baGaaGOmaaaacqGHsislcaaIYaGaciiBaiaac6gacaaIYaGaeyypa0 % JaaGymaiabgkHiTiaaikdaciGGSbGaaiOBaiaaikdacqGHRaWkcaaI % YaGaciiBaiaac6gadaqadaqaaiaaigdacqGHRaWkdaGcaaqaaiaaio % daaSqabaaakiaawIcacaGLPaaaaeaacqGHshI3daGabaabaeqabaGa % amyyaiabg2da9iaaigdaaeaacaWGIbGaeyypa0JaeyOeI0IaaGOmaa % qaaiaadogacqGH9aqpcaaIYaaaaiaawUhaaiabgkDiElaadggacaWG % IbGaam4yaiabg2da9iaaigdacaGGUaGaaiikaiabgkHiTiaaikdaca % GGPaGaaiOlaiaaikdacqGH9aqpcqGHsislcaaI0aaaaaa!8005! \begin{array}{l} = \frac{{{t^2}}}{2} + 2\ln \left| {t + 1} \right|\left| \begin{array}{l} ^{\sqrt 3 }\\ _1 \end{array} \right. = \frac{3}{2} + 2\ln \left( {\sqrt 3 + 1} \right) - \frac{1}{2} - 2\ln 2 = 1 - 2\ln 2 + 2\ln \left( {1 + \sqrt 3 } \right)\\ \Rightarrow \left\{ \begin{array}{l} a = 1\\ b = - 2\\ c = 2 \end{array} \right. \Rightarrow abc = 1.( - 2).2 = - 4 \end{array}\)

Gọi \(\Delta\) là đường thẳng cần tìm

Giả sử \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2 % da9iabfs5aejabgMIihlaadsgacqGHshI3caWGbbWaaeWaaeaacqGH % sislcaaIXaGaeyOeI0IaaGOmaiaadshacaGG7aGaamiDaiaacUdacq % GHsislcaaIXaGaey4kaSIaaG4maiaadshaaiaawIcacaGLPaaaaaa!4B5A! A = \Delta \cap d \Rightarrow A\left( { - 1 - 2t;t; - 1 + 3t} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqaiabg2 % da9iabfs5aejabgMIihlaadsgacaGGNaGaeyO0H4TaamOqamaabmaa % baGaaGOmaiabgUcaRiaadshacaGGNaGaai4oaiabgkHiTiaaigdacq % GHRaWkcaaIYaGaamiDaiaacEcacaGG7aGaeyOeI0IaaGOmaiaadsha % caGGNaaacaGLOaGaayzkaaaaaa!4DFD! B = \Delta \cap d' \Rightarrow B\left( {2 + t'; - 1 + 2t'; - 2t'} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aa8 % HaaeaacaWGbbGaamOqaaGaay51GaGaeyypa0ZaaeWaaeaacaaIYaGa % amiDaiabgUcaRiaadshacaGGNaGaey4kaSIaaG4maiaacUdacqGHsi % slcaWG0bGaey4kaSIaaGOmaiaadshacaGGNaGaeyOeI0IaaGymaiaa % cUdacqGHsislcaaIZaGaamiDaiabgkHiTiaaikdacaWG0bGaai4jai % abgUcaRiaaigdaaiaawIcacaGLPaaaaaa!53D6! \Rightarrow \overrightarrow {AB} = \left( {2t + t' + 3; - t + 2t' - 1; - 3t - 2t' + 1} \right)\) là 1 VTCP của \(\Delta\)

(P) nhận \(\overrightarrow n \left( {1;1;1} \right)\) là 1 VTCP

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaey % yPI41aaeWaaeaacaWGqbaacaGLOaGaayzkaaGaeyO0H49aa8Haaeaa % caWGbbGaamOqaaGaay51Gaaaaa!4107! \Delta \bot \left( P \right) \Rightarrow \overrightarrow {AB} \) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGUbaacaGLxdcaaaa!389B! \overrightarrow n \) là 2 vec tơ cùng phương.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3caaIYaGaamiDaiabgUcaRiaadshacaGGNaGaey4kaSIaaG4maiab % g2da9iabgkHiTiaadshacqGHRaWkcaaIYaGaamiDaiaacEcacqGHRa % WkcaaIXaGaeyypa0JaeyOeI0IaaG4maiaadshacqGHsislcaaIYaGa % amiDaiaacEcacqGHRaWkcaaIXaGaeyi1HS9aaiqaaqaabeqaaiaaio % dacaWG0bGaeyOeI0IaamiDaiaacEcacqGHRaWkcaaI0aGaeyypa0Ja % aGimaaqaaiaaikdacaWG0bGaey4kaSIaaGinaiaadshacaGGNaGaey % OeI0IaaGOmaiabg2da9iaaicdaaaGaay5EaaGaeyi1HS9aaiqaaqaa % beqaaiaadshacqGH9aqpcqGHsislcaaIXaaabaGaamiDaiaacEcacq % GH9aqpcaaIXaaaaiaawUhaaaqaaiabgkDiElaadgeadaqadaqaaiaa % igdacaGG7aGaeyOeI0IaaGymaiaacUdacqGHsislcaaI0aaacaGLOa % GaayzkaaGaaiilaiaadkeadaqadaqaaiaaiodacaGG7aGaaGymaiaa % cUdacqGHsislcaaIYaaacaGLOaGaayzkaaGaeyO0H49aa8Haaeaaca % WGbbGaamOqaaGaay51GaGaeyypa0ZaaeWaaeaacaaIYaGaai4oaiaa % ikdacaGG7aGaaGOmaaGaayjkaiaawMcaaiaac+cacaGGVaWaaeWaae % aacaaIXaGaai4oaiaaigdacaGG7aGaaGymaaGaayjkaiaawMcaaaaa % aa!90F3! \begin{array}{l} \Rightarrow 2t + t' + 3 = - t + 2t' + 1 = - 3t - 2t' + 1 \Leftrightarrow \left\{ \begin{array}{l} 3t - t' + 4 = 0\\ 2t + 4t' - 2 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} t = - 1\\ t' = 1 \end{array} \right.\\ \Rightarrow A\left( {1; - 1; - 4} \right),B\left( {3;1; - 2} \right) \Rightarrow \overrightarrow {AB} = \left( {2;2;2} \right)//\left( {1;1;1} \right) \end{array}\)

Vậy phương trình đường thẳng  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaai % OoamaalaaabaGaamiEaiabgkHiTiaaiodaaeaacaaIXaaaaiabg2da % 9maalaaabaGaamyEaiabgkHiTiaaigdaaeaacaaIXaaaaiabg2da9m % aalaaabaGaamOEaiabgUcaRiaaikdaaeaacaaIXaaaaaaa!446F! \Delta :\frac{{x - 3}}{1} = \frac{{y - 1}}{1} = \frac{{z + 2}}{1}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaaiodacqGH9aqpcaWGTbGaamyzamaaCaaaleqabaGaamiEaaaa % kiabgsDiBlaad2gacqGH9aqpdaWcaaqaaiaadIhacqGHRaWkcaaIZa % aabaGaamyzamaaCaaaleqabaGaamiEaaaaaaGccqGH9aqpcaWGMbWa % aeWaaeaacaWG4baacaGLOaGaayzkaaWaaeWaaeaacaGGQaaacaGLOa % GaayzkaaWaaeWaaeaacaWGebGaam4BaiaaykW7caaMc8UaaGPaVlaa % dwgadaahaaWcbeqaaiaadIhaaaGccqGH+aGpcaaIWaGaeyiaIiIaam % iEaiabgIGiolabl2riHcGaayjkaiaawMcaaaaa!5CF9! x + 3 = m{e^x} \Leftrightarrow m = \frac{{x + 3}}{{{e^x}}} = f\left( x \right)\left( * \right)\left( {Do\,\,\,{e^x} > 0\forall x \in R } \right)\)

Để phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaaiodacqGH9aqpcaWGTbGaamyzamaaCaaaleqabaGaamiEaaaa % aaa!3C9C! x + 3 = m{e^x}\) có 2 nghiệm phân biệt thì phương trình (*) có 2 nghiệm phân biệt.

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiEaiab % gUcaRiaaiodaaeaacaWGLbWaaWbaaSqabeaacaWG4baaaaaaaaa!3F2B! f\left( x \right) = \frac{{x + 3}}{{{e^x}}}\) ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiaa % dwgadaahaaWcbeqaaiaadIhaaaGccqGHsisldaqadaqaaiaadIhacq % GHRaWkcaaIZaaacaGLOaGaayzkaaGaamyzamaaCaaaleqabaGaamiE % aaaaaOqaaiaadwgadaahaaWcbeqaaiaaikdacaWG4baaaaaakiabg2 % da9maalaaabaGaeyOeI0IaamiEaiabgkHiTiaaikdaaeaacaWGLbWa % aWbaaSqabeaacaWG4baaaaaakiabg2da9iaaicdacqGHuhY2caWG4b % Gaeyypa0JaeyOeI0IaaGOmaaaa!55DD! f'\left( x \right) = \frac{{{e^x} - \left( {x + 3} \right){e^x}}}{{{e^{2x}}}} = \frac{{ - x - 2}}{{{e^x}}} = 0 \Leftrightarrow x = - 2\)

Số nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaaaaa!3B5D! m = f\left( x \right)\) là số giao điểm của đồ thị hàm số y = m và y =f(x)

Dựa vào BBT ta có phương trình (*) có 2 nghiệm phân biệt  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4TaaG % imaiabgYda8iaad2gacqGH8aapcaWGLbWaaWbaaSqabeaacaaIYaaa % aaaa!3DD8! \Rightarrow 0 < m < {e^2}\)

Mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgI % GiolablssiIkabgkDiElaad2gacqGHiiIZdaGadaqaaiaaigdacaGG % 7aGaaGOmaiaacUdacaaIZaGaai4oaiaaisdacaGG7aGaaGynaiaacU % dacaaI2aGaai4oaiaaiEdaaiaawUhacaGL9baaaaa!4A92! m \in Z \Rightarrow m \in \left\{ {1;2;3;4;5;6;7} \right\}\)

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaWGMbGaai4jamaabmaabaGaamiEaiabgkHiTiaaigda % aiaawIcacaGLPaaacqGHRaWkcaaIYaGaamiEaiabgkHiTiaaikdacq % GH9aqpcaaIWaGaeyi1HSTaamOzaiaacEcadaqadaqaaiaadIhacqGH % sislcaaIXaaacaGLOaGaayzkaaGaey4kaSIaaGOmamaabmaabaGaam % iEaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH9aqpcaaIWaaaaa!5417! y' = f'\left( {x - 1} \right) + 2x - 2 = 0 \Leftrightarrow f'\left( {x - 1} \right) + 2\left( {x - 1} \right) = 0\)

Đặt t = x - 1 ta có  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacaWGMbGaai4jamaabmaabaGaamiDaaGaayjkaiaawMcaaiabgUca % RiaaikdacaWG0bGaeyypa0JaaGimaiabgsDiBlaadAgacaGGNaWaae % WaaeaacaWG0baacaGLOaGaayzkaaGaeyOeI0YaaeWaaeaacqGHsisl % caaIYaGaamiDaaGaayjkaiaawMcaaiabg2da9iaaicdaaaa!4D58! y'f'\left( t \right) + 2t = 0 \Leftrightarrow f'\left( t \right) - \left( { - 2t} \right) = 0\)

Vẽ đồ thị hàm số y = f'(t) và y = -2t trên cùng mặt phẳng tọa độ ta có:

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGHLjYScaaIWaGaeyi1HSTaamOzaiaacEcadaqadaqaaiaadsha % aiaawIcacaGLPaaacqGHLjYScqGHsislcaaIYaGaamiDaiabgkDiEd % aa!4756! y' \ge 0 \Leftrightarrow f'\left( t \right) \ge - 2t \Rightarrow \) Đồ thị hàm số y = f'(t) nằm trên đường thẳng y = -2t

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaGaeyO0 % H4TaamiDaiabgIGiopaabmaabaGaaGimaiaacUdacaaIXaaacaGLOa % GaayzkaaGaeyO0H4naaa!4728! x \in \left( {1;2} \right) \Rightarrow t \in \left( {0;1} \right) \Rightarrow \) thỏa mãn.

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaeyOeI0IaaGymaiaacUdacaaIWaaacaGLOaGaayzk % aaGaeyO0H4TaamiDaiabgIGiopaabmaabaGaeyOeI0IaaGOmaiaacU % dacqGHsislcaaIXaaacaGLOaGaayzkaaGaeyO0H4naaa!49EF! x \in \left( { - 1;0} \right) \Rightarrow t \in \left( { - 2; - 1} \right) \Rightarrow \) không thỏa mãn.

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGimaiaacUdacaaIXaaacaGLOaGaayzkaaGaeyO0 % H4TaamiDaiabgIGiopaabmaabaGaeyOeI0IaaGymaiaacUdacaaIWa % aacaGLOaGaayzkaaGaeyO0H4naaa!4813! x \in \left( {0;1} \right) \Rightarrow t \in \left( { - 1;0} \right) \Rightarrow \) không thỏa mãn.

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaeyOeI0IaaGOmaiaacUdacqGHsislcaaIXaaacaGL % OaGaayzkaaGaeyO0H4TaamiDaiabgIGiopaabmaabaGaeyOeI0IaaG % 4maiaacUdacqGHsislcaaIYaaacaGLOaGaayzkaaGaeyO0H4naaa!4AE0! x \in \left( { - 2; - 1} \right) \Rightarrow t \in \left( { - 3; - 2} \right) \Rightarrow \) không thỏa mãn.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaciiBaiaac6gadaqadaqaaiaadIhacqGHRaWkcaaI1aaa % caGLOaGaayzkaaaaaiabgUcaRmaalaaabaGaaGymaaqaaiaaiodada % ahaaWcbeqaaiaadIhaaaGccqGHsislcaaIXaaaaiabg2da9iaadIha % cqGHRaWkcaWGHbGaeyi1HSTaamOzamaabmaabaGaamiEaaGaayjkai % aawMcaaiabg2da9maalaaabaGaaGymaaqaaiGacYgacaGGUbWaaeWa % aeaacaWG4bGaey4kaSIaaGynaaGaayjkaiaawMcaaaaacqGHRaWkda % WcaaqaaiaaigdaaeaacaaIZaWaaWbaaSqabeaacaWG4baaaOGaeyOe % I0IaaGymaaaacqGHsislcaWG4bGaeyypa0JaamyyamaabmaabaGaai % OkaaGaayjkaiaawMcaaaaa!5ED7! \frac{1}{{\ln \left( {x + 5} \right)}} + \frac{1}{{{3^x} - 1}} = x + a \Leftrightarrow f\left( x \right) = \frac{1}{{\ln \left( {x + 5} \right)}} + \frac{1}{{{3^x} - 1}} - x = a\left( * \right)\)

Xét hàm số : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqa % aiGacYgacaGGUbWaaeWaaeaacaWG4bGaey4kaSIaaGynaaGaayjkai % aawMcaaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaIZaWaaWbaaSqa % beaacaWG4baaaOGaeyOeI0IaaGymaaaacqGHsislcaWG4baaaa!4871! f\left( x \right) = \frac{1}{{\ln \left( {x + 5} \right)}} + \frac{1}{{{3^x} - 1}} - x\)

ĐKXĐ : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGHRaWkcaaI1aGaeyOpa4JaaGimaaqaaiGacYgacaGG % UbWaaeWaaeaacaWG4bGaey4kaSIaaGynaaGaayjkaiaawMcaaiabgc % Mi5kaaicdaaeaacaaIZaWaaWbaaSqabeaacaWG4baaaOGaeyOeI0Ia % aGymaiabgcMi5kaaicdaaaGaay5EaaGaeyi1HS9aaiqaaqaabeqaai % aadIhacqGH+aGpcqGHsislcaaI1aaabaGaamiEaiabgUcaRiaaiwda % cqGHGjsUcaaIXaaabaGaaG4mamaaCaaaleqabaGaamiEaaaakiabgc % Mi5kaaigdaaaGaay5EaaGaeyi1HS9aaiqaaqaabeqaaiaadIhacqGH % +aGpcqGHsislcaaI1aaabaGaamiEaiabgcMi5kabgkHiTiaaisdaae % aacaWG4bGaeyiyIKRaaGimaaaacaGL7baaaaa!69FA! \left\{ \begin{array}{l} x + 5 > 0\\ \ln \left( {x + 5} \right) \ne 0\\ {3^x} - 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x > - 5\\ x + 5 \ne 1\\ {3^x} \ne 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x > - 5\\ x \ne - 4\\ x \ne 0 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % iraiabg2da9maabmaabaGaeyOeI0IaaGynaiaacUdacqGHsislcaaI % 0aaacaGLOaGaayzkaaGaeyOkIG8aaeWaaeaacqGHsislcaaI0aGaai % 4oaiaaicdaaiaawIcacaGLPaaacqGHQicYdaqadaqaaiaaicdacaGG % 7aGaey4kaSIaeyOhIukacaGLOaGaayzkaaaaaa!4D01! \Rightarrow D = \left( { - 5; - 4} \right) \cup \left( { - 4;0} \right) \cup \left( {0; + \infty } \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiab % gkHiTiaaigdaaeaaciGGSbGaaiOBamaaCaaaleqabaGaaGOmaaaakm % aabmaabaGaamiEaiabgUcaRiaaiwdaaiaawIcacaGLPaaaaaGaeyOe % I0YaaSaaaeaacaaIZaWaaWbaaSqabeaacaWG4baaaaGcbaWaaeWaae % aacaaIZaWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaaGymaaGaayjk % aiaawMcaamaaCaaaleqabaGaaGOmaaaaaaGccqGHsislcaaIXaGaey % ipaWJaaGimaiabgcGiIiaadIhacqGHiiIZcaWGebaaaa!544F! f'\left( x \right) = \frac{{ - 1}}{{{{\ln }^2}\left( {x + 5} \right)}} - \frac{{{3^x}}}{{{{\left( {{3^x} - 1} \right)}^2}}} - 1 < 0\forall x \in D\)

BBT : 

Từ BBT suy ra phương trình (*) có 2 nghiệm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yyaiabgwMiZkaaisdaaaa!3BBA! \Leftrightarrow a \ge 4\)

Kết hợp ĐK \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yyaiabgIGiopaacmaabaGaaGinaiaacUdacaGGUaGaaiOlaiaac6ca % caGG7aGaaGOmaiaaicdacaaIXaGaaGioaaGaay5Eaiaaw2haaaaa!4431! \Rightarrow a \in \left\{ {4;...;2018} \right\}\). Vậy có 2015 giá trị của a thỏa mãn.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WG4bGaamOzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWG % KbGaamiEaiabg2da9maapehabaGaamiEaiaadsgadaqadaqaaiaadA % gadaqadaqaaiaadIhaaiaawIcacaGLPaaaaiaawIcacaGLPaaacqGH % 9aqpcaWG4bGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaamaaee % aaeaqabeaadaahaaWcbeqaaiaaikdaaaaakeaadaWgaaWcbaGaaGim % aaqabaaaaOGaay5bSdGaeyOeI0Yaa8qCaeaacaWGMbWaaeWaaeaaca % WG4baacaGLOaGaayzkaaGaamizaiaadIhacqGH9aqpcaaIYaGaamOz % amaabmaabaGaaGOmaaGaayjkaiaawMcaaiabgkHiTmaapehabaGaam % OzamaabmaabaGaamiEaaGaayjkaiaawMcaaiaadsgacaWG4baaleaa % caaIWaaabaGaaGOmaaqdcqGHRiI8aaWcbaGaaGimaaqaaiaaikdaa0 % Gaey4kIipaaSqaaiaaicdaaeaacaaIYaaaniabgUIiYdaaleaacaaI % WaaabaGaaGOmaaqdcqGHRiI8aaaa!6EC1! \int\limits_0^2 {xf'\left( x \right)dx = \int\limits_0^2 {xd\left( {f\left( x \right)} \right) = xf\left( x \right)\left| \begin{array}{l} ^2\\ _0 \end{array} \right. - \int\limits_0^2 {f\left( x \right)dx = 2f\left( 2 \right) - \int\limits_0^2 {f\left( x \right)dx} } } } \)

Theo bài ra ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGMb % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaey4kaSIaamOzamaabmaa % baGaaGOmaiabgkHiTiaadIhaaiaawIcacaGLPaaacqGH9aqpcaWG4b % WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaadIhacqGHRaWk % caaIYaGaeyiaIiIaamiEaiabgIGiolabl2riHkabgkDiElaadAgada % qadaqaaiaaicdaaiaawIcacaGLPaaacqGHRaWkcaWGMbWaaeWaaeaa % caaIYaaacaGLOaGaayzkaaGaeyypa0JaaGOmaiabgkDiElaadAgada % qadaqaaiaaikdaaiaawIcacaGLPaaacqGH9aqpcaaIYaGaeyOeI0Ia % amOzamaabmaabaGaaGimaaGaayjkaiaawMcaaiabg2da9iabgkHiTi % aaigdaaeaacqGHshI3daWdXbqaaiaadIhacaWGMbGaai4jamaabmaa % baGaamiEaaGaayjkaiaawMcaaiaadsgacaWG4bGaeyypa0JaeyOeI0 % IaaGOmaiabgkHiTmaapehabaGaamOzamaabmaabaGaamiEaaGaayjk % aiaawMcaaiaadsgacaWG4bGaeyypa0JaeyOeI0IaaGOmaiabgkHiTm % aapehabaGaamOzamaabmaabaGaamiDaaGaayjkaiaawMcaaiaadsga % caWG0baaleaacaaIWaaabaGaaGOmaaqdcqGHRiI8aaWcbaGaaGimaa % qaaiaaikdaa0Gaey4kIipaaSqaaiaaicdaaeaacaaIYaaaniabgUIi % Ydaaaaa!8BB1! \begin{array}{l} f\left( x \right) + f\left( {2 - x} \right) = {x^2} - 2x + 2\forall x \in R \Rightarrow f\left( 0 \right) + f\left( 2 \right) = 2 \Rightarrow f\left( 2 \right) = 2 - f\left( 0 \right) = - 1\\ \Rightarrow \int\limits_0^2 {xf'\left( x \right)dx = - 2 - \int\limits_0^2 {f\left( x \right)dx = - 2 - \int\limits_0^2 {f\left( t \right)dt} } } \end{array}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaikdacqGHsislcaWG4bGaeyO0H4TaamizaiaadshacqGH9aqp % cqGHsislcaWGKbGaamiEaaaa!42B1! t = 2 - x \Rightarrow dt = - dx\). Đổi cận  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaaIWaGaeyO0H4TaamiDaiabg2da9iaaikda % aeaacaWG4bGaeyypa0JaaGOmaiabgkDiElaadshacqGH9aqpcaaIWa % aaaiaawUhaaaaa!46BF! \left\{ \begin{array}{l} x = 0 \Rightarrow t = 2\\ x = 2 \Rightarrow t = 0 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aa8 % qCaeaacaWGMbWaaeWaaeaacaWG0baacaGLOaGaayzkaaGaamizaiaa % dIhacqGH9aqpcqGHsisldaWdXbqaaiaadAgadaqadaqaaiaaikdacq % GHsislcaWG4baacaGLOaGaayzkaaGaamizaiaadIhacqGH9aqpdaWd % XbqaaiaadAgadaqadaqaaiaaikdacqGHsislcaWG4baacaGLOaGaay % zkaaGaamizaiaadIhaaSqaaiaaicdaaeaacaaIYaaaniabgUIiYdaa % leaacaaIYaaabaGaaGimaaqdcqGHRiI8aaWcbaGaaGimaaqaaiaaik % daa0Gaey4kIipaaaa!5A3A! \Rightarrow \int\limits_0^2 {f\left( t \right)dx = - \int\limits_2^0 {f\left( {2 - x} \right)dx = \int\limits_0^2 {f\left( {2 - x} \right)dx} } } \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3daWdXbqaaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWG % KbGaamiEaiabg2da9maapehabaGaamOzamaabmaabaGaaGOmaiabgk % HiTiaadIhaaiaawIcacaGLPaaaaSqaaiaaicdaaeaacaaIYaaaniab % gUIiYdaaleaacaaIWaaabaGaaGOmaaqdcqGHRiI8aOGaamizaiaadI % haaeaacqGHshI3caaIYaWaa8qCaeaacaWGMbWaaeWaaeaacaWG4baa % caGLOaGaayzkaaGaamizaiaadIhacqGH9aqpdaWdXbqaaiaadAgada % qadaqaaiaadIhaaiaawIcacaGLPaaacaWGKbGaamiEaiabgUcaRmaa % pehabaGaamOzamaabmaabaGaaGOmaiabgkHiTiaadIhaaiaawIcaca % GLPaaacaWGKbGaamiEaaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIipa % aSqaaiaaicdaaeaacaaIYaaaniabgUIiYdaaleaacaaIWaaabaGaaG % OmaaqdcqGHRiI8aaGcbaGaeyO0H4TaaGOmamaapehabaGaamOzamaa % bmaabaGaamiEaaGaayjkaiaawMcaaiaadsgacaWG4bGaeyypa0Zaa8 % qCaeaadaWadaqaaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaa % cqGHRaWkcaWGMbWaaeWaaeaacaaIYaGaeyOeI0IaamiEaaGaayjkai % aawMcaaaGaay5waiaaw2faaiaadsgacaWG4baaleaacaaIWaaabaGa % aGOmaaqdcqGHRiI8aaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIipaaO % qaaiabgkDiElaaikdadaWdXbqaaiaadAgadaqadaqaaiaadIhaaiaa % wIcacaGLPaaacaWGKbGaamiEaiabg2da9maapehabaWaaeWaaeaaca % WG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaadIhacqGH % RaWkcaaIYaaacaGLOaGaayzkaaGaamizaiaadIhaaSqaaiaaicdaae % aacaaIYaaaniabgUIiYdaaleaacaaIWaaabaGaaGOmaaqdcqGHRiI8 % aaGcbaGaeyO0H4TaaGOmamaapehabaGaamOzamaabmaabaGaamiEaa % GaayjkaiaawMcaaiaadsgacaWG4bGaeyypa0ZaaeWaaeaadaWcaaqa % aiaadIhadaahaaWcbeqaaiaaiodaaaaakeaacaaIZaaaaiabgkHiTi % aadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamiEaaGa % ayjkaiaawMcaamaaeeaaeaqabeaadaahaaWcbeqaaiaaikdaaaaake % aadaWgaaWcbaGaaGimaaqabaaaaOGaay5bSdaaleaacaaIWaaabaGa % aGOmaaqdcqGHRiI8aOGaeyypa0ZaaSaaaeaacaaI4aaabaGaaG4maa % aaaeaacqGHshI3daWdXbqaaiaadAgadaqadaqaaiaadIhaaiaawIca % caGLPaaacaWGKbGaamiEaiabg2da9maalaaabaGaaGinaaqaaiaaio % daaaaaleaacaaIWaaabaGaaGOmaaqdcqGHRiI8aaaaaa!D48F! \begin{array}{l} \Rightarrow \int\limits_0^2 {f\left( x \right)dx = \int\limits_0^2 {f\left( {2 - x} \right)} } dx\\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx = \int\limits_0^2 {f\left( x \right)dx + \int\limits_0^2 {f\left( {2 - x} \right)dx} } } \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx = \int\limits_0^2 {\left[ {f\left( x \right) + f\left( {2 - x} \right)} \right]dx} } \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx = \int\limits_0^2 {\left( {{x^2} - 2x + 2} \right)dx} } \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx = \left( {\frac{{{x^3}}}{3} - {x^2} + 2x} \right)\left| \begin{array}{l} ^2\\ _0 \end{array} \right.} = \frac{8}{3}\\ \Rightarrow \int\limits_0^2 {f\left( x \right)dx = \frac{4}{3}} \end{array}\)

Vậy  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WG4bGaamOzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWG % KbGaamiEaiabg2da9iabgkHiTiaaikdacqGHsisldaWcaaqaaiaais % daaeaacaaIZaaaaiabg2da9iabgkHiTmaalaaabaGaaGymaiaaicda % aeaacaaIZaaaaaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIipaaaa!4A2E! \int\limits_0^2 {xf'\left( x \right)dx = - 2 - \frac{4}{3} = - \frac{{10}}{3}} \)

Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maaemaabaWaaSaaaeaa % caWG4baabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaig % daaaGaeyOeI0IaamyBaaGaay5bSlaawIa7aaaa!4406! f\left( x \right) = \left| {\frac{x}{{{x^2} + 1}} - m} \right|\) có TXĐ: D = R

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiEaaqa % aiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaaaaiabgk % HiTiaad2gaaaa!40E5! g\left( x \right) = \frac{x}{{{x^2} + 1}} - m\) ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zaiaacE % cadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiaa % dIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaGaeyOeI0Iaam % iEaiaac6cacaaIYaGaamiEaaqaamaabmaabaGaamiEamaaCaaaleqa % baGaaGOmaaaakiabgUcaRiaaigdaaiaawIcacaGLPaaadaahaaWcbe % qaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaacqGHsislcaWG4bWaaWba % aSqabeaacaaIYaaaaOGaey4kaSIaaGymaaqaamaabmaabaGaamiEam % aaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigdaaiaawIcacaGLPaaa % daahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaaGimaiabgsDiBlaadI % hacqGH9aqpcqGHXcqScaaIXaaaaa!5D73! g'\left( x \right) = \frac{{{x^2} + 1 - x.2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{ - {x^2} + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0 \Leftrightarrow x = \pm 1\)

 Hàm số y = g(x) có 2 điểm cực trị.

Xét phương trình hoành độ giao điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigda % aaGaeyOeI0IaamyBaiabg2da9iaaicdacqGHuhY2daWcaaqaaiaadI % hacqGHsislcaWGTbWaaeWaaeaacaWG4bWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaaGymaaGaayjkaiaawMcaaaqaaiaadIhadaahaaWcbe % qaaiaaikdaaaGccqGHRaWkcaaIXaaaaiabg2da9iaaicdacqGHuhY2 % cqGHsislcaWGTbGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRi % aadIhacqGHsislcaWGTbGaeyypa0JaaGimaaaa!5981! \frac{x}{{{x^2} + 1}} - m = 0 \Leftrightarrow \frac{{x - m\left( {{x^2} + 1} \right)}}{{{x^2} + 1}} = 0 \Leftrightarrow - m{x^2} + x - m = 0\), phương trình có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaey % ypa0JaaGymaiabgkHiTiaaisdacaWGTbWaaWbaaSqabeaacaaIYaaa % aaaa!3CA1! \Delta = 1 - 4{m^2}\) chưa xác định dấu nên có tối đa 2 nghiệm.

Vậy hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maaemaabaWaaSaaaeaa % caWG4baabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaig % daaaGaeyOeI0IaamyBaaGaay5bSlaawIa7aaaa!4406! f\left( x \right) = \left| {\frac{x}{{{x^2} + 1}} - m} \right|\) có tối đa 2 + 2 = 4 cực trị.

Đặc biệt hóa, coi ABCD.A'B'C'D' là khối lập phương cạnh bằng 1 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OvamaaBaaaleaacaWGbbGaamOqaiaadoeacaWGebGaaiOlaiaadgea % caGGNaGaamOqaiaacEcacaWGdbGaai4jaiaadseacaGGNaaabeaaki % abg2da9iaaigdacqGH9aqpcaWGwbaaaa!469E! \Rightarrow {V_{ABCD.A'B'C'D'}} = 1 = V\)

Dễ thấy MNPQEF là khối bát diện đều cạnh \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyuaiaadw % eacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadkeacaWGebGa % eyypa0ZaaSaaaeaadaGcaaqaaiaaikdaaSqabaaakeaacaaIYaaaaa % aa!3E64! QE = \frac{1}{2}BD = \frac{{\sqrt 2 }}{2}\)

Vậy  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGnbGaamOtaiaadcfacaWGrbGaamyraiaadAeaaeqaaOGa % eyypa0ZaaSaaaeaadaqadaqaamaalaaabaWaaOaaaeaacaaIYaaale % qaaaGcbaGaaGOmaaaaaiaawIcacaGLPaaadaahaaWcbeqaaiaaioda % aaGcdaGcaaqaaiaaikdaaSqabaaakeaacaaIZaaaaiabg2da9maala % aabaGaaGymaaqaaiaaiAdaaaGaeyypa0ZaaSaaaeaacaWGwbaabaGa % aGOnaaaaaaa!480A! {V_{MNPQEF}} = \frac{{{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^3}\sqrt 2 }}{3} = \frac{1}{6} = \frac{V}{6}\)

Gắn hệ trục tọa độ như hình vẽ.

Diện tích phần tô đậm là  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9iaaisdadaWadaqaamaapehabaWaaeWaaeaadaGcaaqaaiaaikda % caWG4baaleqaaOGaeyOeI0IaaGimaaGaayjkaiaawMcaaiaadsgaca % WG4baaleaacaaIWaaabaGaaGymaaqdcqGHRiI8aOGaey4kaSYaa8qC % aeaadaqadaqaamaakaaabaGaaGOmaiaadIhaaSqabaGccqGHsislca % aIYaWaaOaaaeaadaqadaqaaiaadIhacqGHsislcaaIXaaacaGLOaGa % ayzkaaWaaWbaaSqabeaacaaIZaaaaaqabaaakiaawIcacaGLPaaaca % WGKbGaamiEaaWcbaGaaGymaaqaaiaaikdaa0Gaey4kIipaaOGaay5w % aiaaw2faaiabg2da9maalaaabaGaaGymaiaaigdacaaIYaaabaGaaG % ymaiaaiwdaaaWaaeWaaeaacaWGKbGaamyBamaaCaaaleqabaGaaGOm % aaaaaOGaayjkaiaawMcaaiabgIKi7kaaiEdacaaI0aGaaG4namaabm % aabaGaam4yaiaad2gadaahaaWcbeqaaiaaikdaaaaakiaawIcacaGL % Paaaaaa!67B2! S = 4\left[ {\int\limits_0^1 {\left( {\sqrt {2x} - 0} \right)dx} + \int\limits_1^2 {\left( {\sqrt {2x} - 2\sqrt {{{\left( {x - 1} \right)}^3}} } \right)dx} } \right] = \frac{{112}}{{15}}\left( {d{m^2}} \right) \approx 747\left( {c{m^2}} \right)\)

Theo bài ra ta có:

+) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6baacaGLhWUaayjcSdGaeyypa0JaaGOmaiabgkDiEdaa!3E34! \left| z \right| = 2 \Rightarrow \) Tập hợp các điểm biểu diễn số phức z là đường tròn tâm \(I_1(0;0)\) bán kính \(R_1 = 2\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WGPbaacaGLhWUaayjcSdWaaqWaaeaacaqG3bGaeyOeI0YaaSaaaeaa % caaIYaGaeyOeI0IaaGynaiaadMgaaeaacaWGPbaaaaGaay5bSlaawI % a7aiabg2da9iaaigdacqGHuhY2daabdaqaaiaabEhacqGHsisldaqa % daqaaiabgkHiTiaaiwdacqGHsislcaaIYaGaamyAaaGaayjkaiaawM % caaaGaay5bSlaawIa7aiabg2da9iaaigdaaaa!5413! \left| i \right|\left| {{\rm{w}} - \frac{{2 - 5i}}{i}} \right| = 1 \Leftrightarrow \left| {{\rm{w}} - \left( { - 5 - 2i} \right)} \right| = 1\)

 Tập hợp các điểm biểu diễn số phức w là đường tròn tâm \(I_2(-5;-2)\) bán kính  \(R_2=1\)

Đặt  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9maaemaabaGaamOEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaa % bEhacaWG6bGaeyOeI0IaaGinaaGaay5bSlaawIa7aiabg2da9maaem % aabaGaamOEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaabEhacaWG % 6bGaeyOeI0IaamOEaiaac6cadaqdaaqaaiaadQhaaaaacaGLhWUaay % jcSdGaeyypa0ZaaqWaaeaacaWG6baacaGLhWUaayjcSdWaaqWaaeaa % caWG6bGaeyOeI0Iaae4DaiabgkHiTmaanaaabaGaamOEaaaaaiaawE % a7caGLiWoacqGH9aqpcaaIYaWaaqWaaeaacaWG6bGaeyOeI0Iaae4D % aiabgkHiTmaanaaabaGaamOEaaaaaiaawEa7caGLiWoaaaa!6519! T = \left| {{z^2} - {\rm{w}}z - 4} \right| = \left| {{z^2} - {\rm{w}}z - z.\overline z } \right| = \left| z \right|\left| {z - {\rm{w}} - \overline z } \right| = 2\left| {z - {\rm{w}} - \overline z } \right|\)

Đặt  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEaiabg2 % da9iaadggacqGHRaWkcaWGIbGaamyAaiaacYcadaqadaqaaiaadgga % caGGSaGaamOyaiabgIGiolabl2riHcGaayjkaiaawMcaaaaa!4340! z = a + bi,\left( {a,b \in R} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WG6baaaiabg2da9iaadggacqGHsislcaWGIbGaamyAaaaa!3BB2! \overline z = a - bi\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OEaiabgkHiTmaanaaabaGaamOEaaaacqGH9aqpcaaIYaGaamOyaiaa % dMgaaaa!3EE4! \Rightarrow z - \overline z = 2bi\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ivaiabg2da9iaaikdadaabdaqaaiaaikdacaWGIbGaamyAaiabgkHi % TiaadEhaaiaawEa7caGLiWoaaaa!4288! \Rightarrow T = 2\left| {2bi - w} \right|\)

Gọi M(0;2b) là điểm biểu diễn số phức 2bi, N là điểm biểu diễn số phức w.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ivaiabg2da9iaaikdacaWGnbGaamOtamaaBaaaleaaciGGTbGaaiyA % aiaac6gaaeqaaOGaeyi1HSTaamytaiaad6eadaWgaaWcbaGaciyBai % aacMgacaGGUbaabeaaaaa!4698! \Rightarrow T = 2M{N_{\min }} \Leftrightarrow M{N_{\min }}\)

Do  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6baacaGLhWUaayjcSdGaeyypa0JaaGOmaiabgkDiElaadggadaah % aaWcbeqaaiaaikdaaaGccqGHRaWkcaWGIbWaaWbaaSqabeaacaaIYa % aaaOGaeyypa0JaaGinaiabgsDiBlabgkHiTiaaikdacqGHKjYOcaWG % IbGaeyizImQaaGOmaiabgsDiBlabgkHiTiaaisdacqGHKjYOcaaIYa % GaamOyaiabgsMiJkaaisdaaaa!5771! \left| z \right| = 2 \Rightarrow {a^2} + {b^2} = 4 \Leftrightarrow - 2 \le b \le 2 \Leftrightarrow - 4 \le 2b \le 4\)

 Tập hợp các điểm M là đoạn AB với  A(-4 ; 0) B(4 ; 0)

Dựa vào hình vẽ ta thấy  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiaad6 % eadaWgaaWcbaGaciyBaiaacMgacaGGUbaabeaakiabg2da9iaaisda % cqGHuhY2caWGnbWaaeWaaeaacqGHsislcaaI0aGaai4oaiabgkHiTi % aaikdaaiaawIcacaGLPaaacaGGSaGaamOtamaabmaabaGaaGimaiaa % cUdacqGHsislcaaIYaaacaGLOaGaayzkaaaaaa!4B5D! M{N_{\min }} = 4 \Leftrightarrow M\left( { - 4; - 2} \right),N\left( {0; - 2} \right)\)

Vậy  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa % aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0JaaGOmaiaac6ca % caaI0aGaeyypa0JaaGioaaaa!3ECF! {T_{\min }} = 2.4 = 8\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGMb % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyOpa4Jaci4CaiaacMga % caGGUbWaaSaaaeaacqaHapaCcaWG4baabaGaaGOmaaaacqGHRaWkca % WGTbGaeyiaIiIaamiEaiabgIGiopaadmaabaGaeyOeI0IaaGymaiaa % cUdacaaIZaaacaGLBbGaayzxaaGaeyi1HSTaam4zamaabmaabaGaam % iEaaGaayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaadIhaaiaa % wIcacaGLPaaacqGHsislciGGZbGaaiyAaiaac6gadaWcaaqaaiabec % 8aWjaadIhaaeaacaaIYaaaaiabg6da+iaad2gacqGHaiIicaWG4bGa % eyicI48aamWaaeaacqGHsislcaaIXaGaai4oaiaaiodaaiaawUfaca % GLDbaaaeaacqGHshI3caWGTbGaeyipaWZaaCbeaeaaciGGTbGaaiyA % aiaac6gaaSqaamaadmaabaGaeyOeI0IaaGymaiaacUdacaaIZaaaca % GLBbGaayzxaaaabeaakiaadEgadaqadaqaaiaadIhaaiaawIcacaGL % Paaaaaaa!76EE! \begin{array}{l} f\left( x \right) > \sin \frac{{\pi x}}{2} + m\forall x \in \left[ { - 1;3} \right] \Leftrightarrow g\left( x \right) = f\left( x \right) - \sin \frac{{\pi x}}{2} > m\forall x \in \left[ { - 1;3} \right]\\ \Rightarrow m < \mathop {\min }\limits_{\left[ { - 1;3} \right]} g\left( x \right) \end{array}\)

Từ đồ thị hàm số y = f'(x) ta suy ra BBT đồ thị hàm số y = f(x) như sau:

Dựa vào BBT ta thấy  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabgwMiZkaadAgadaqadaqaaiaa % igdaaiaawIcacaGLPaaacqGHaiIicaWG4bGaeyicI48aamWaaeaacq % GHsislcaaIXaGaai4oaiaaiodaaiaawUfacaGLDbaaaaa!46C1! f\left( x \right) \ge f\left( 1 \right)\forall x \in \left[ { - 1;3} \right]\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG4b % GaeyicI48aamWaaeaacqGHsislcaaIXaGaai4oaiaaiodaaiaawUfa % caGLDbaacqGHshI3daWcaaqaaiabec8aWjaadIhaaeaacaaIYaaaai % abgIGiopaadmaabaGaeyOeI0YaaSaaaeaacqaHapaCaeaacaaIYaaa % aiaacUdadaWcaaqaaiaaiodacqaHapaCaeaacaaIYaaaaaGaay5wai % aaw2faaiabgkDiElabgkHiTiaaigdacqGHKjYOciGGZbGaaiyAaiaa % c6gadaWcaaqaaiabec8aWjaadIhaaeaacaaIYaaaaiabgsMiJkaaig % daaeaacqGHuhY2cqGHsislcaaIXaGaeyizImQaeyOeI0Iaci4Caiaa % cMgacaGGUbWaaSaaaeaacqaHapaCcaWG4baabaGaaGOmaaaacqGHKj % YOcaaIXaaabaGaeyO0H4TaamOzamaabmaabaGaaGymaaGaayjkaiaa % wMcaaiabgkHiTiaaigdacqGHKjYOcaWGMbWaaeWaaeaacaWG4baaca % GLOaGaayzkaaGaeyOeI0Iaci4CaiaacMgacaGGUbWaaSaaaeaacqaH % apaCcaWG4baabaGaaGOmaaaacqGHuhY2caWGNbWaaeWaaeaacaWG4b % aacaGLOaGaayzkaaGaeyyzImRaamOzamaabmaabaGaaGymaaGaayjk % aiaawMcaaiabgkHiTiaaigdacqGHshI3daWfqaqaaiGac2gacaGGPb % GaaiOBaaWcbaWaamWaaeaacqGHsislcaaIXaGaai4oaiaaiodaaiaa % wUfacaGLDbaaaeqaaOGaam4zamaabmaabaGaamiEaaGaayjkaiaawM % caaiabg2da9iaadAgadaqadaqaaiaaigdaaiaawIcacaGLPaaacqGH % sislcaaIXaaaaaa!A04A! \begin{array}{l} x \in \left[ { - 1;3} \right] \Rightarrow \frac{{\pi x}}{2} \in \left[ { - \frac{\pi }{2};\frac{{3\pi }}{2}} \right] \Rightarrow - 1 \le \sin \frac{{\pi x}}{2} \le 1\\ \Leftrightarrow - 1 \le - \sin \frac{{\pi x}}{2} \le 1\\ \Rightarrow f\left( 1 \right) - 1 \le f\left( x \right) - \sin \frac{{\pi x}}{2} \Leftrightarrow g\left( x \right) \ge f\left( 1 \right) - 1 \Rightarrow \mathop {\min }\limits_{\left[ { - 1;3} \right]} g\left( x \right) = f\left( 1 \right) - 1 \end{array}\)

Vậy  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgY % da8iaadAgadaqadaqaaiaaigdaaiaawIcacaGLPaaacqGHsislcaaI % Xaaaaa!3CC1! m < f\left( 1 \right) - 1\)

Gọi (P) là mặt phẳng đi qua B và vuông góc với  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabgk % DiEpaabmaabaGaamiuaaGaayjkaiaawMcaaiaacQdacaaIYaGaamiE % aiabgUcaRiaadMhacqGHRaWkcaWG6bGaeyOeI0IaaGymaiabg2da9i % aaicdacaGGUaGaaGPaVlaaykW7caaMc8oaaa!4A8B! d \Rightarrow \left( P \right):2x + y + z - 1 = 0.\,\,\,\)

\(\Delta\) đi qua B và vuông góc với  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabgk % DiElabfs5aejabgkOimpaabmaabaGaamiuaaGaayjkaiaawMcaaaaa % !3EFA! d \Rightarrow \Delta \subset \left( P \right)\)

Gọi H, K lần lượt là hình chiếu của A lên (P) và  ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadI % eacqGHKjYOcaWGbbGaam4saaaa!3AD2! AH \le AK\)

Do đó để khoảng cách từ A đến \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % isaiabgIGiolabfs5aebaa!3C08! \Rightarrow H \in \Delta \)  là nhỏ nhất  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % isaiabgIGiolabfs5aebaa!3C08! \Rightarrow H \in \Delta \)

Phương trình AH đi qua A và nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1bWaaSbaaSqaaiaadsgaaeqaaaGccaGLxdcacqGH9aqpdaqadaqa % aiaaikdacaGG7aGaaGymaiaacUdacaaIXaaacaGLOaGaayzkaaaaaa!4000! \overrightarrow {{u_d}} = \left( {2;1;1} \right)\) là 1 VTCP là  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaaI2aGaey4kaSIaaGOmaiaadshaaeaacaWG % 5bGaeyypa0JaaG4maiabgUcaRiaadshaaeaacaWG6bGaeyypa0Jaey % OeI0IaaGOmaiabgUcaRiaadshaaaGaay5Eaaaaaa!4695! \left\{ \begin{array}{l} x = 6 + 2t\\ y = 3 + t\\ z = - 2 + t \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGib % GaeyicI4SaamyqaiaadIeacqGHshI3caWGibWaaeWaaeaacaaI2aGa % ey4kaSIaaGOmaiaadshacaGG7aGaaG4maiabgUcaRiaadshacaGG7a % GaeyOeI0IaaGOmaiabgUcaRiaadshaaiaawIcacaGLPaaaaeaacaWG % ibGaeyicI48aaeWaaeaacaWGqbaacaGLOaGaayzkaaGaeyO0H4TaaG % OmamaabmaabaGaaGOnaiabgUcaRiaaikdacaWG0baacaGLOaGaayzk % aaGaey4kaSIaaG4maiabgUcaRiaadshacqGHsislcaaIYaGaey4kaS % IaamiDaiabgkHiTiaaigdacqGH9aqpcaaIWaGaeyi1HSTaaGOnaiaa % dshacqGHRaWkcaaIXaGaaGOmaiabg2da9iaaicdacqGHuhY2caWG0b % Gaeyypa0JaeyOeI0IaaGOmaaqaaiabgkDiElaadIeadaqadaqaaiaa % ikdacaGG7aGaaGymaiaacUdacqGHsislcaaI0aaacaGLOaGaayzkaa % aaaaa!7817! \begin{array}{l} H \in AH \Rightarrow H\left( {6 + 2t;3 + t; - 2 + t} \right)\\ H \in \left( P \right) \Rightarrow 2\left( {6 + 2t} \right) + 3 + t - 2 + t - 1 = 0 \Leftrightarrow 6t + 12 = 0 \Leftrightarrow t = - 2\\ \Rightarrow H\left( {2;1; - 4} \right) \end{array}\)

đi qua B, H nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGcbGaamisaaGaay51GaWaaeWaaeaacaaIXaGaai4oaiaaigdacaGG % 7aGaeyOeI0IaaG4maaGaayjkaiaawMcaaaaa!3F63! \overrightarrow {BH} \left( {1;1; - 3} \right)\) là 1 VTCP.

Gọi H, K lần lượt là hình chiếu của A lên (P) và d ta có , khi đó mặt phẳng (P) chứa đường thẳng d thỏa mãn khoảng cách từ điểm A đến (P) lớn nhất ; (P) nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGbbGaam4saaGaay51Gaaaaa!393E! \overrightarrow {AK} \) là 1 VTPT.

Gọi  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaabm % aabaGaaGymaiabgUcaRiaaikdacaWG0bGaai4oaiabgkHiTiaaikda % cqGHRaWkcaWG0bGaai4oaiaaikdacaWG0baacaGLOaGaayzkaaGaey % icI4SaamizaiabgkDiEpaaFiaabaGaamyqaiaadUeaaiaawEniaiab % g2da9maabmaabaGaaGOmaiaadshacqGHsislcaaIXaGaai4oaiaads % hacqGHRaWkcaaIXaGaai4oaiaaikdacaWG0bGaeyOeI0IaaGinaaGa % ayjkaiaawMcaaaaa!57CA! K\left( {1 + 2t; - 2 + t;2t} \right) \in d \Rightarrow \overrightarrow {AK} = \left( {2t - 1;t + 1;2t - 4} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1bWaaSbaaSqaaiaadsgaaeqaaaGccaGLxdcadaqadaqaaiaaikda % caGG7aGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaaaaa!3EFB! \overrightarrow {{u_d}} \left( {2;1;2} \right)\) là 1 VTCP của d

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3daWhcaqaaiaadgeacaWGlbaacaGLxdcacaGGUaWaa8HaaeaacaWG % 1bWaaSbaaSqaaiaadsgaaeqaaaGccaGLxdcacqGH9aqpcaaIWaGaey % i1HSTaaGinaiaadshacqGHsislcaaIYaGaey4kaSIaamiDaiabgUca % RiaaigdacqGHRaWkcaaI0aGaamiDaiabgkHiTiaaiIdacqGH9aqpca % aIWaGaeyi1HSTaaGyoaiaadshacqGHsislcaaI5aGaeyypa0JaaGim % aiabgsDiBlaadshacqGH9aqpcaaIXaaabaGaeyO0H4Taam4samaabm % aabaGaaG4maiaacUdacqGHsislcaaIXaGaai4oaiaaikdaaiaawIca % caGLPaaacqGHshI3daWhcaqaaiaadgeacaWGlbaacaGLxdcacqGH9a % qpdaqadaqaaiaaigdacaGG7aGaaGOmaiaacUdacqGHsislcaaIYaaa % caGLOaGaayzkaaaabaGaeyO0H49aaeWaaeaacaWGqbaacaGLOaGaay % zkaaGaaiOoaiaadIhacqGHsislcaaIZaGaey4kaSIaaGOmamaabmaa % baGaamyEaiabgUcaRiaaigdaaiaawIcacaGLPaaacqGHsislcaaIYa % WaaeWaaeaacaWG6bGaeyOeI0IaaGOmaaGaayjkaiaawMcaaiabg2da % 9iaaicdacqGHuhY2caWG4bGaey4kaSIaaGOmaiaadMhacqGHsislca % aIYaGaamOEaiabgUcaRiaaiodacqGH9aqpcaaIWaaaaaa!955D! \begin{array}{l} \Rightarrow \overrightarrow {AK} .\overrightarrow {{u_d}} = 0 \Leftrightarrow 4t - 2 + t + 1 + 4t - 8 = 0 \Leftrightarrow 9t - 9 = 0 \Leftrightarrow t = 1\\ \Rightarrow K\left( {3; - 1;2} \right) \Rightarrow \overrightarrow {AK} = \left( {1;2; - 2} \right)\\ \Rightarrow \left( P \right):x - 3 + 2\left( {y + 1} \right) - 2\left( {z - 2} \right) = 0 \Leftrightarrow x + 2y - 2z + 3 = 0 \end{array}\)

Mặt cầu \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGtbaacaGLOaGaayzkaaGaaiOoamaabmaabaGaamiEaiabgkHiTiaa % iodaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda % qadaqaaiaadMhacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSYaaeWaaeaacaWG6bGaey4kaSIaaGymaa % GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabg2da9iaaikda % caaIWaaaaa!4CB1! \left( S \right):{\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z + 1} \right)^2} = 20\) có tâm I ( 3;2;-1), bán kính  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9maakaaabaGaaGOmaiaaicdaaSqabaGccqGH9aqpcaaIYaWaaOaa % aeaacaaI1aaaleqaaaaa!3C08! R = \sqrt {20} = 2\sqrt 5 \)

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabg2 % da9iaadsgadaqadaqaaiaadMeacaGG7aWaaeWaaeaacaWGqbaacaGL % OaGaayzkaaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaadaabdaqaai % aaiodacqGHRaWkcaaIYaGaaiOlaiaaikdacqGHsislcaaIYaWaaeWa % aeaacqGHsislcaaIXaaacaGLOaGaayzkaaGaey4kaSIaaG4maaGaay % 5bSlaawIa7aaqaamaakaaabaGaaGymaiabgUcaRiaaisdacqGHRaWk % caaI0aaaleqaaaaakiabg2da9maalaaabaGaaGymaiaaikdaaeaaca % aIZaaaaiabg2da9iaaisdaaaa!55E8! d = d\left( {I;\left( P \right)} \right) = \frac{{\left| {3 + 2.2 - 2\left( { - 1} \right) + 3} \right|}}{{\sqrt {1 + 4 + 4} }} = \frac{{12}}{3} = 4\)

Gọi r là đường kính đường tròn giao tuyến của (P) và (S) ta có:

  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuamaaCa % aaleqabaGaaGOmaaaakiabg2da9iaadsgadaahaaWcbeqaaiaaikda % aaGccqGHRaWkcaWGYbWaaWbaaSqabeaacaaIYaaaaOGaeyi1HSTaam % OCaiabg2da9maakaaabaGaamOuamaaCaaaleqabaGaaGOmaaaakiab % gkHiTiaadsgadaahaaWcbeqaaiaaikdaaaaabeaakiabg2da9maaka % aabaGaaGOmaiaaicdacqGHsislcaaIXaGaaGOnaaWcbeaakiabg2da % 9iaaikdaaaa!4D33! {R^2} = {d^2} + {r^2} \Leftrightarrow r = \sqrt {{R^2} - {d^2}} = \sqrt {20 - 16} = 2\)