Giá trị của \(A_8^3\) là 336

\(\overrightarrow a = 2\overrightarrow j + 3\overrightarrow k =2(0;1;0)+3(0;0;1)=(0;2;3)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabg2da % 9iaaikdacaGGOaGaaGimaiaacUdacaaIXaGaai4oaiaaicdacaGGPa % Gaey4kaSIaaG4maiaacIcacaaIWaGaai4oaiaaicdacaGG7aGaaGym % aiaacMcaaeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca % aMc8UaaGPaVlabg2da9iaacIcacaaIWaGaai4oaiaaikdacaGG7aGa % aG4maiaacMcaaaaa!6227! \begin{array}{l} \,\,\,\, = 2(0;1;0) + 3(0;0;1)\\ \,\,\,\, = (0;2;3) \end{array}\)

Một vectơ pháp tuyến của mặt phẳng \((P):2x-y+1=0\) là 

\(\vec{n}=(2;-1;0)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaaiikaiaaiIdacaWGHbGa % aiykaiabg2da9iGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabe % aakiaaiIdacqGHRaWkciGGSbGaai4BaiaacEgadaWgaaWcbaGaaGOm % aaqabaGccaWGHbGaeyypa0JaaG4maiabgUcaRiGacYgacaGGVbGaai % 4zamaaBaaaleaacaaIYaaabeaakiaadggaaaa!4F17! {\log _2}(8a) = {\log _2}8 + {\log _2}a = 3 + {\log _2}a\)

Với \(t=0\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaaIYaaabaGaamyEaiabg2da9iabgkHiTiaa % igdaaeaacaWG6bGaeyypa0JaaGymaaaacaGL7baaaaa!4040! \left\{ \begin{array}{l} x = 2\\ y = - 1\\ z = 1 \end{array} \right.\).

Vậy điểm \(M(2;-1;1)\) là điểm thuộc đường thẳng \(d:\left\{ \begin{array}{l} x = 2 + t\\ y = - 1 + t\\ z = 1 - 2t \end{array} \right.\)

Phương trình mặt cầu tâm \(I(1;-2;1)\) và bán kính bằng 2 là

\((x-1)^2+(y+2)^2+(z-1)^2=2^2\)

\((x-1)^2+(y+2)^2+(z-1)^2=4\)

\(\int {\frac{1}{{{{\cos }^2}2x}}dx} =\frac{1}{2}tan2x+C\)

TXĐ: \(D=\mathbb{R}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaaIYa % WaaWbaaSqabeaacaWG4bGaey4kaSIaaGymaaaakiabg2da9iaaigda % caaI2aaabaGaeyi1HSTaaGOmamaaCaaaleqabaGaamiEaiabgUcaRi % aaigdaaaGccqGH9aqpcaaIYaWaaWbaaSqabeaacaaI0aaaaaGcbaGa % eyi1HSTaamiEaiabgUcaRiaaigdacqGH9aqpcaaI0aaabaGaeyi1HS % TaamiEaiabg2da9iaaiodaaaaa!5080! \begin{array}{l} \,\,\,\,\,\,\,{2^{x + 1}} = 16\\ \Leftrightarrow {2^{x + 1}} = {2^4}\\ \Leftrightarrow x + 1 = 4\\ \Leftrightarrow x = 3 \end{array}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaWG4bGaaiykaiabgUcaRiaaigdacqGH9aqpcaaIWaGaeyi1HSTa % amOzaiaacIcacaWG4bGaaiykaiabg2da9iabgkHiTiaaigdaaaa!44DC! f(x) + 1 = 0 \Leftrightarrow f(x) = - 1\)

Số nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaWG4bGaaiykaiabgUcaRiaaigdacqGH9aqpcaaIWaGaeyi1HSTa % amOzaiaacIcacaWG4bGaaiykaiabg2da9iabgkHiTiaaigdaaaa!44DC! f(x) + 1 = 0 \) chính là số giao điểm của đồ thị hàm số \(y=f(x)\) và \(y=-1\).

Từ bảng biến thiên suy ra số nghiệm của phương trình là 3

Quan sát đồ thị hàm số ta thấy giá trị cực đại của hàm số là 3

Số các số tự nhiên có 3 chữa số khác nhau được lập thành từ các chữ số \(0;1;2;3;4;5;6;7\) là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaaDa % aaleaacaaI4aaabaGaaGOmaaaakiabgkHiTiaadgeadaqhaaWcbaGa % aG4naaqaaiaaikdaaaGccqGH9aqpcaaIYaGaaGyoaiaaisdaaaa!3F18! A_8^2 - A_7^2 = 294\)

Đồ thị là đồ thị của hàm số bậc 3 nên loại câu B, C

 Nhìn vào đồ thị ta thấy \( \mathop {\lim }\limits_{x \to + \infty } y = - \infty \) nên chọn câu D

Thể tích khối trụ có bán kính đáy và chiều cao bằng 2 là:\(V=\pi R^2h=\pi.4.2=8\pi\)

\((u_n)\) là cấp số cộng nên 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG1b % WaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaamyDamaaBaaaleaacaaI % XaaabeaakiabgUcaRiaadsgaaeaacqGHshI3caWGKbGaeyypa0Jaam % yDamaaBaaaleaacaaIYaaabeaakiabgkHiTiaadwhadaWgaaWcbaGa % aGymaaqabaGccqGH9aqpcaaIZaGaeyOeI0IaaGymaiabg2da9iaaik % daaaaa!4ADF! \begin{array}{l} {u_2} = {u_1} + d\\ \Rightarrow d = {u_2} - {u_1} = 3 - 1 = 2 \end{array}\)

\(\int\limits_{-2}^{ 2} {f(x)dx} =\int\limits_{-2}^{ 0} {f(x)dx} +\int\limits_{0}^{ 2} {f(x)dx} \)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\int\limits_{0}^{- 2} {f(x)dx}+\int\limits_{0}^{ 2} {f(x)dx} \)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-17+4=-13\)

Từ bảng biến thiên ta có

\(f'(x)>0\Leftrightarrow x\in(1;3)\)

Vậy hàm số đã cho đồng biến trên \((1;3)\)

Bán kính đáy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCaiabg2 % da9iaadIgacaGGUaGaciiDaiaacggacaGGUbGaaGOnaiaaicdadaah % aaWcbeqaaiaaicdaaaGccqGH9aqpcaaIYaWaaOaaaeaacaaIZaaale % qaaaaa!4165! r = h.\tan {30^0} = 2\frac{1}{\sqrt {3}} =\frac{2\sqrt{3}}{3}\)

ĐK: \((x-3)>0\Leftrightarrow x>3\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGSb % Gaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGccaGGOaGaamiEaiab % gkHiTiaaiodacaGGPaGaeyypa0JaaG4maaqaaiabgsDiBlaadIhacq % GHsislcaaIZaGaeyypa0JaaGOmamaaCaaaleqabaGaaG4maaaaaOqa % aiabgsDiBlaadIhacqGHsislcaaIZaGaeyypa0JaaGioaaqaaiabgs % DiBlaadIhacqGH9aqpcaaIXaGaaGymaaaaaa!53DA! \begin{array}{l} \,\,\,\,\,\,\,\,{\log _2}(x - 3) = 3\\ \Leftrightarrow x - 3 = {2^3}\\ \Leftrightarrow x - 3 = 8\\ \Leftrightarrow x = 11\,\mathrm{(nhận)} \end{array}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6baacaGLhWUaayjcSdGaeyypa0ZaaOaaaeaacaaI0aWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSIaaiikaiabgkHiTiaaiodacaGGPaWaaW % baaSqabeaacaaIYaaaaaqabaGccqGH9aqpcaaI1aaaaa!4378! \left| z \right| = \sqrt {{4^2} + {{( - 3)}^2}} = 5\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaacM % gacaGGTbWaaSaaaeaacaaIYaGaamOBaiabgUcaRiaaiodaaeaacaWG % UbGaey4kaSIaaGymaaaacqGH9aqpciGGSbGaaiyAaiaac2gadaWcaa % qaaiaaikdacqGHRaWkdaWcaaqaaiaaiodaaeaacaWGUbaaaaqaaiaa % igdacqGHRaWkdaWcaaqaaiaaigdaaeaacaWGUbaaaaaacqGH9aqpca % aIYaaaaa!4B12! \lim \frac{{2n + 3}}{{n + 1}} = \lim \frac{{2 + \frac{3}{n}}}{{1 + \frac{1}{n}}} = 2\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGbbGaai4jaiaac6cacaWGbbGaamOqaiaadoeaaeqaaOGa % eyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGwbWaaSbaaSqaai % aadgeacaGGNaGaamyqaiaadkeacaWGdbGaamiraaqabaGccqGH9aqp % daWcaaqaaiaaigdaaeaacaaIYaaaaiaac6cadaWcaaqaaiaaigdaae % aacaaIZaaaaiaadAfadaWgaaWcbaGaamyqaiaadkeacaWGdbGaamir % aiaac+cacaWGbbGaai4jaiaadkeacaGGNaGaam4qaiaacEcacaWGeb % Gaai4jaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaI2aaaaiaa % dggadaahaaWcbeqaaiaaiodaaaaaaa!577D! {V_{A'.ABC}} = \frac{1}{2}{V_{A'ABCD}} = \frac{1}{2}.\frac{1}{3}{V_{ABCD.A'B'C'D'}} = \frac{1}{6}{a^3}\)

Nhìn vào đồ thị hàm số ta thấy khi đi qua 4 điểm \(x=-1; x=0; x=1; x=2\) thì \(f'(x)\) đổi dấu

Vậy hàm số có 4 điểm cực trị.

Mặt cầu có tâm \(I(2;-2;0)\)

Bán kính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9maakaaabaGaaGOmamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaa % cIcacqGHsislcaaIYaGaaiykamaaCaaaleqabaGaaGOmaaaakiabgU % caRiaaicdadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aaaleqa % aOGaeyypa0JaaGimaaaa!4475! R = \sqrt {{2^2} + {{( - 2)}^2} + {0^2} - 4} = 2\)

BC là giao tuyến của \((ABCD)\) và \((A'D'BC)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadkeacaWGdbGaeyyPI4LaamyqaiaacEcacaWGcbGaeyOGIWSa % aiikaiaadgeacaGGNaGaamiraiaacEcacaWGcbGaam4qaiaacMcaae % aacaWGcbGaam4qaiabgwQiEjaadgeacaWGcbGaeyOGIWSaaiikaiaa % dgeacaWGcbGaam4qaiaadseacaGGPaaaaiaawUhaaaaa!4F95! \left\{ \begin{array}{l} BC \bot A'B \subset (A'D'BC)\\ BC \bot AB \subset (ABCD) \end{array} \right.\)

Vậy góc giữa hai mặt phẳng \((ABCD)\) và \((A'D'BC)\) là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaqiaa % qaaiaadgeacaGGNaGaamOqaiaadgeaaiaawkWaaaqaaaaaaa!39BA! \begin{array}{l} \widehat {A'BA}\\ \end{array}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGG0b % Gaaiyyaiaac6gadaqiaaqaaiaadgeacaGGNaGaamOqaiaadgeaaiaa % wkWaaiabg2da9maalaaabaGaamyqaiaadgeacaGGNaaabaGaamyqai % aadkeaaaGaeyypa0JaaGymaaqaamaaHaaabaGaamyqaiaacEcacaWG % cbGaamyqaaGaayPadaGaeyypa0JaaGinaiaaiwdadaahaaWcbeqaai % aaicdaaaaaaaa!4A50! \begin{array}{l} \tan \widehat {A'BA} = \frac{{AA'}}{{AB}} = 1\\ \Rightarrow\widehat {A'BA} = {45^0} \end{array}\)

Vậy góc giữa hai mặt phẳng là \(45^0\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG5b % Gaeyypa0ZaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4k % aSIaamiEaiabgUcaRiaaisdaaeaacaWG4bGaey4kaSIaaGymaaaacq % GHshI3caWG5bGaai4jaiabg2da9maalaaabaGaamiEamaaCaaaleqa % baGaaGOmaaaakiabgUcaRiaaikdacaWG4bGaeyOeI0IaaG4maaqaai % aacIcacaWG4bGaey4kaSIaaGymaiaacMcadaahaaWcbeqaaiaaikda % aaaaaaGcbaGaamyEaiaacEcacqGH9aqpcaaIWaGaeyi1HS9aaSaaae % aacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadIha % cqGHsislcaaIZaaabaGaaiikaiaadIhacqGHRaWkcaaIXaGaaiykam % aaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpcaaIWaGaeyi1HS9aamqa % aqaabeqaaiaadIhacqGH9aqpcaaIXaaabaGaamiEaiabg2da9iabgk % HiTiaaiodaaaGaay5waaaabaGaamiEaiabgIGiolaabUfacaaIWaGa % ai4oaiaaikdacaGGDbGaeyO0H4TaamiEaiabg2da9iaaigdaaeaaca % WG5bGaaiikaiaaicdacaGGPaGaeyypa0JaaGinaaqaaiaadMhacaGG % OaGaaGymaiaacMcacqGH9aqpcaaIZaaabaGaamyEaiaacIcacaaIYa % Gaaiykaiabg2da9maalaaabaGaaGymaiaaicdaaeaacaaIZaaaaaaa % aa!8860! \begin{array}{l} y = \frac{{{x^2} + x + 4}}{{x + 1}} \Rightarrow y' = \frac{{{x^2} + 2x - 3}}{{{{(x + 1)}^2}}}\\ \mathrm{Cho}\,y' = 0 \Leftrightarrow \frac{{{x^2} + 2x - 3}}{{{{(x + 1)}^2}}} = 0 \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - 3 \end{array} \right.\\ \mathrm{Do}\,x \in {\rm{[}}0;2] \Rightarrow x = 1\\ y(0) = 4\\ y(1) = 3\\ y(2) = \frac{{10}}{3} \end{array}\)

Vậy giá trị nhỏ nhất của hàm số trên đoạn \([0;2]\) là 3

Mặt phẳng vuông góc với AB nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGbbGaamOqaaGaay51GaGaeyypa0JaaiikaiaaikdacaGG7aGaaGym % aiaacUdacqGHsislcaaIXaGaaiykaaaa!4030! \overrightarrow {AB} = (2;1; - 1)\) là một vectơ pháp tuyển của mặt phẳng

Phương trình mặt phẳng qua A và vuông góc với AB:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaaIYa % GaaiikaiaadIhacqGHsislcaaIWaGaaiykaiabgUcaRiaacIcacaWG % 5bGaeyOeI0IaaGymaiaacMcacqGHsislcaGGOaGaamOEaiabgkHiTi % aaikdacaGGPaGaeyypa0JaaGimaaqaaiabgsDiBlaaikdacaWG4bGa % ey4kaSIaamyEaiabgkHiTiaadQhacqGHRaWkcaaIXaGaeyypa0JaaG % imaaaaaa!5180! \begin{array}{l} 2(x - 0) + (y - 1) - (z - 2) = 0\\ \Leftrightarrow 2x + y - z + 1 = 0 \end{array}\)

ĐK:\(x>0\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGSb % Gaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGccaWG4bGaey4kaSIa % ciiBaiaac+gacaGGNbWaaSbaaSqaaiaaiodaaeqaaOGaamiEaiabg2 % da9iaaigdacqGHRaWkciGGSbGaai4BaiaacEgadaWgaaWcbaGaaGOm % aaqabaGccaWG4bGaaiOlaiGacYgacaGGVbGaai4zamaaBaaaleaaca % aIZaaabeaakiaadIhaaeaacqGHuhY2ciGGSbGaai4BaiaacEgadaWg % aaWcbaGaaGOmaaqabaGccaWG4bGaeyOeI0IaaGymaiabg2da9iGacY % gacaGGVbGaai4zamaaBaaaleaacaaIZaaabeaakiaadIhacaGGUaGa % aiikaiGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakiaadI % hacqGHsislcaaIXaGaaiykaaqaaiabgsDiBlaacIcaciGGSbGaai4B % aiaacEgadaWgaaWcbaGaaGOmaaqabaGccaWG4bGaeyOeI0IaaGymai % aacMcacqGHsislciGGSbGaai4BaiaacEgadaWgaaWcbaGaaG4maaqa % baGccaWG4bGaaiOlaiaacIcaciGGSbGaai4BaiaacEgadaWgaaWcba % GaaGOmaaqabaGccaWG4bGaeyOeI0IaaGymaiaacMcacqGH9aqpcaaI % WaaabaGaeyi1HSTaaiikaiGacYgacaGGVbGaai4zamaaBaaaleaaca % aIYaaabeaakiaadIhacqGHsislcaaIXaGaaiykamaabmaabaGaaGym % aiabgkHiTiGacYgacaGGVbGaai4zamaaBaaaleaacaaIZaaabeaaki % aadIhaaiaawIcacaGLPaaacqGH9aqpcaaIWaaabaGaeyi1HS9aamqa % aqaabeqaaiGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaaki % aadIhacqGHsislcaaIXaGaeyypa0JaaGimaaqaaiaaigdacqGHsisl % ciGGSbGaai4BaiaacEgadaWgaaWcbaGaaG4maaqabaGccaWG4bGaey % ypa0JaaGimaaaacaGLBbaaaeaacqGHuhY2daWabaabaeqabaGaciiB % aiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaamiEaiabg2da9i % aaigdaaeaaciGGSbGaai4BaiaacEgadaWgaaWcbaGaaG4maaqabaGc % caWG4bGaeyypa0JaaGymaaaacaGLBbaaaeaacqGHuhY2daWabaabae % qabaGaamiEaiabg2da9iaaikdaaeaacaWG4bGaeyypa0JaaG4maaaa % caGLBbaaaaaa!BEE0! \begin{array}{l} {\log _2}x + {\log _3}x = 1 + {\log _2}x.{\log _3}x\\ \Leftrightarrow {\log _2}x - 1 = {\log _3}x.({\log _2}x - 1)\\ \Leftrightarrow ({\log _2}x - 1) - {\log _3}x.({\log _2}x - 1) = 0\\ \Leftrightarrow ({\log _2}x - 1)\left( {1 - {{\log }_3}x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} {\log _2}x - 1 = 0\\ 1 - {\log _3}x = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} {\log _2}x = 1\\ {\log _3}x = 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 3 \end{array} \right. \end{array}\)

Vậy \(x_1^2+x_2^2=13\)

\(|z|=|(1-2i)(1+i)^2|=|1-2i|.|(1+i)^2|=\sqrt{5}.|1+i|^2=2\sqrt{5}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG6b % Gaeyypa0JaaG4maiabgkHiTiaaikdacaWGPbGaeyO0H49aa0aaaeaa % caWG6baaaiabg2da9iaaiodacqGHRaWkcaaIYaGaamyAaaqaaiaabE % hacqGH9aqpcaWG6bGaey4kaSIaamyAamaanaaabaGaamOEaaaacqGH % 9aqpcaGGOaGaaG4maiabgkHiTiaaikdacaWGPbGaaiykaiabgUcaRi % aadMgacaGGOaGaaG4maiabgUcaRiaaikdacaWGPbGaaiykaiabg2da % 9iaaigdacqGHRaWkcaWGPbaaaaa!58A4! \begin{array}{l} z = 3 - 2i \Rightarrow \overline z = 3 + 2i\\ {\rm{w}} = z + i\overline z = (3 - 2i) + i(3 + 2i) = 1 + i \end{array}\)

Vậy điểm biểu diễn hình học của số phức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaae4Daaaa!36ED! {\rm{w}}\) là \((1;1)\)

Ta có:

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaqGYa % GaaeOzaiaabIcacaqG4bGaaeykaiaab2cacaqG1aGaaeypaiaabcda % aeaacqGHuhY2caqGMbGaaeikaiaabIhacaqGPaGaaeypamaalaaaba % GaaGynaaqaaiaaikdaaaaaaaa!44A7! \begin{array}{l} {\rm{2f(x) - 5 = 0}}\\ \Leftrightarrow {\rm{f(x) = }}\frac{5}{2} \end{array}\)

Số nghiệm của phương trình \(2f(x)-5=0\) là số giao điểm của đồ thị hàm số \(y=f(x)\) và đồ thị hàm số \(y=\frac{5}{2}\)

do \(1<\frac{5}{2}<3\) nên từ bảng biến thiên suy ra phương trình có 4 nghiệm phân biệt

Gọi \(r\,(r>0)\) là bán kính mặt cầu. Ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGtb % Gaeyypa0JaaGymaiaaiAdacqaHapaCcqGHuhY2caaI0aGaeqiWdaNa % amOCamaaCaaaleqabaGaaGOmaaaakiabg2da9iaaigdacaaI2aGaeq % iWdahabaGaeyO0H4TaamOCaiabg2da9iaaikdaaaaa!4B25! \begin{array}{l} S = 16\pi \Leftrightarrow 4\pi {r^2} = 16\pi \\ \Rightarrow r = 2 \end{array}\)

Thể tích khối cầu:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGinaaqaaiaaiodaaaGaeqiWdaNaamOCamaaCaaa % leqabaGaaG4maaaakiabg2da9maalaaabaGaaGinaaqaaiaaiodaaa % GaeqiWdaNaaGioaiabg2da9maalaaabaGaaG4maiaaikdacqaHapaC % aeaacaaIZaaaaaaa!4720! V = \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi 8 = \frac{{32\pi }}{3}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGpbGaamyqaaGaay51GaGaeyypa0JaaiikaiaaigdacaGG7aGaaGim % aiaacUdacqGHsislcaaIYaGaaiykaaaa!403C! \overrightarrow {OA} = (1;0; - 2)\). Do mặt phẳng vuông góc với OA nên \(\vec{OA}\) là một vectơ pháp tuyến của mặt phẳng

Phương trình mặt phẳng qua A và vuông góc với OA có dạng

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaGGOa % GaamiEaiabgkHiTiaaigdacaGGPaGaey4kaSIaaGimaiaacIcacaWG % 5bGaeyOeI0IaaGimaiaacMcacqGHsislcaaIYaGaaiikaiaadQhacq % GHRaWkcaaIYaGaaiykaiabg2da9iaaicdaaeaacqGHuhY2caWG4bGa % eyOeI0IaaGOmaiaadQhacqGHsislcaaI1aGaeyypa0JaaGimaaaaaa!505E! \begin{array}{l} (x - 1) + 0(y - 0) - 2(z + 2) = 0\\ \Leftrightarrow x - 2z - 5 = 0 \end{array}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaisdacaWG4bGaeyO0H4TaamizaiaadshacqGH9aqpcaaI0aGa % amizaiaadIhacqGHshI3caWGKbGaamiEaiabg2da9maalaaabaGaam % izaiaadshaaeaacaaI0aaaaaaa!498F! t = 4x \Rightarrow dt = 4dx \Rightarrow dx = \frac{{dt}}{4}\)

Đổi cận:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG4b % Gaeyypa0JaaGimaiabgkDiElaadshacqGH9aqpcaaIWaaabaGaamiE % aiabg2da9iaaigdacqGHshI3caWG0bGaeyypa0JaaGinaaaaaa!45A5! \begin{array}{l} x = 0 \Rightarrow t = 0\\ x = 1 \Rightarrow t = 4 \end{array}\)
Khi đó 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WGMbGaaiikaiaaisdacaWG4bGaaiykaiaadsgacaWG4baaleaacaaI % WaaabaGaaGymaaqdcqGHRiI8aOGaeyypa0Zaa8qCaeaacaWGMbGaai % ikaiaadshacaGGPaWaaSaaaeaacaWGKbGaamiDaaqaaiaaisdaaaGa % eyypa0ZaaSaaaeaacaaIXaaabaGaaGinaaaaaSqaaiaaicdaaeaaca % aI0aaaniabgUIiYdGcdaWdXbqaaiaadAgacaGGOaGaamiDaiaacMca % caWGKbGaamiDaaWcbaGaaGimaaqaaiaaisdaa0Gaey4kIipakiabg2 % da9maalaaabaGaeyOeI0IaaGymaaqaaiaaisdaaaaaaa!59B3! \int\limits_0^1 {f(4x)dx} = \int\limits_0^4 {f(t)\frac{{dt}}{4} = \frac{1}{4}} \int\limits_0^4 {f(t)dt} = \frac{{ - 1}}{4}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGtb % Gaamytaiabg2da9maalaaabaGaaGOmaaqaaiaaiodaaaGaam4uaiaa % dkeacqGHshI3daWcaaqaaiaadofacaWGcbaabaGaam4uaiaad2eaaa % Gaeyypa0ZaaSaaaeaacaaIZaaabaGaaGOmaaaaaeaacaWGtbGaamOt % aiabg2da9maalaaabaGaaGinaaqaaiaaiwdaaaGaam4uaiaadoeacq % GHshI3daWcaaqaaiaadofacaWGdbaabaGaam4uaiaad6eaaaGaeyyp % a0ZaaSaaaeaacaaI1aaabaGaaGinaaaaaaaa!5240! \begin{array}{l} SM = \frac{2}{3}SB \Rightarrow \frac{{SB}}{{SM}} = \frac{3}{2}\\ SN = \frac{4}{5}SC \Rightarrow \frac{{SC}}{{SN}} = \frac{5}{4} \end{array}\)

Ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa % qaaiaadAfadaWgaaWcbaGaam4uaiaac6cacaWGbbGaamOqaiaadoea % aeqaaaGcbaGaamOvamaaBaaaleaacaWGtbGaaiOlaiaadgeacaWGnb % GaamOtaaqabaaaaOGaeyypa0ZaaSaaaeaacaWGtbGaamyqaaqaaiaa % dofacaWGbbaaaiaac6cadaWcaaqaaiaadofacaWGcbaabaGaam4uai % aad2eaaaGaaiOlamaalaaabaGaam4uaiaadoeaaeaacaWGtbGaamOt % aaaacqGH9aqpcaaIXaGaaiOlamaalaaabaGaaG4maaqaaiaaikdaaa % GaaiOlamaalaaabaGaaGynaaqaaiaaisdaaaGaeyypa0ZaaSaaaeaa % caaIXaGaaGynaaqaaiaaiIdaaaaabaGaeyO0H4TaamOvamaaBaaale % aacaWGtbGaaiOlaiaadgeacaWGnbGaamOtaaqabaGccqGH9aqpcaWG % wbWaaSbaaSqaaiaadofacaGGUaGaamyqaiaadkeacaWGdbaabeaaki % aacQdadaWcaaqaaiaaigdacaaI1aaabaGaaGioaaaacqGH9aqpcaaI % 3aGaaGimaiaadggadaahaaWcbeqaaiaaiodaaaGccaGG6aWaaSaaae % aacaaIXaGaaGynaaqaaiaaiIdaaaGaeyypa0ZaaSaaaeaacaaIXaGa % aGymaiaaikdacaWGHbWaaWbaaSqabeaacaaIZaaaaaGcbaGaaG4maa % aaaaaa!739E! \begin{array}{l} \frac{{{V_{S.ABC}}}}{{{V_{S.AMN}}}} = \frac{{SA}}{{SA}}.\frac{{SB}}{{SM}}.\frac{{SC}}{{SN}} = 1.\frac{3}{2}.\frac{5}{4} = \frac{{15}}{8}\\ \Rightarrow {V_{S.AMN}} = {V_{S.ABC}}:\frac{{15}}{8} = 70{a^3}:\frac{{15}}{8} = \frac{{112{a^3}}}{3} \end{array}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGMb % GaaiikaiaadIhacaGGPaGaeyypa0JaamiEamaaCaaaleqabaGaaG4m % aaaakiabgkHiTiaaiodacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey % 4kaSIaaGOmaiabgkDiElaadAgacaGGNaGaaiikaiaadIhacaGGPaGa % eyypa0JaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislca % aI2aGaamiEaaqaaiaadAgacaGGNaGaaiikaiaadIhacaGGPaGaeyyp % a0JaaGimaiabgsDiBlaaiodacaWG4bWaaWbaaSqabeaacaaIYaaaaO % GaeyOeI0IaaGOnaiaadIhacqGH9aqpcaaIWaGaeyi1HS9aamqaaqaa % beqaaiaadIhacqGH9aqpcaaIWaaabaGaamiEaiabg2da9iaaigdaae % aacaWG4bGaeyypa0JaeyOeI0IaaGymaaaacaGLBbaaaaaa!69B4! \begin{array}{l} f(x) = {x^3} - 3{x^2} + 2 \Rightarrow f'(x) = 3{x^2} - 6x\\ f'(x) = 0 \Leftrightarrow 3{x^2} - 6x = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 2 \end{array} \right. \end{array}\)

Để hàm số đồng biến thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEaiaacMcacqGH+aGpcaaIWaGaeyi1HS9aamqaaqaa % beqaaiaadIhacqGH8aapcaaIWaaabaGaamiEaiabg6da+iaaikdaaa % Gaay5waaaaaa!447B! f'(x) > 0 \Leftrightarrow \left[ \begin{array}{l} x < 0\\ x > 2 \end{array} \right.\)

Vậy hàm số đồng biến trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiabgk % HiTiabg6HiLkaacUdacaaIWaGaaiykaiabgQIiilaacIcacaaIYaGa % ai4oaiabgUcaRiabg6HiLkaacMcaaaa!41EA! ( - \infty ;0) \cup (2; + \infty )\)

ĐK: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaaI0a % GaeyOeI0IaamiEamaaCaaaleqabaGaaGOmaaaakiabg6da+iaaicda % aeaacqGHsislcaaIYaGaeyipaWJaamiEaiabgYda8iaaikdaaaaa!40C1! \begin{array}{l} 4 - {x^2} > 0\\ \Leftrightarrow - 2 < x < 2 \end{array}\)

Vậy tập xác định của hàm số là \((-2;2)\)

Ta có: 

\(\int_{ - 2}^2 {f\left( x \right)dx} = \int_{ - 2}^1 {f\left( x \right)} dx + \int_1^2 {f\left( x \right)dx} \\ = \int_{ - 2}^1 {\left| {f\left( x \right)} \right|} dx - \int_1^2 {\left| {f\left( x \right)} \right|dx} = {S_1} - {S_2}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa % qaaiGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakiaadkga % aeaaciGGSbGaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGccaWGHb % aaaiabg2da9maalaaabaGaaGOmaiaadogaaeaacaaIYaaaaaqaaiab % gsDiBlGacYgacaGGVbGaai4zamaaBaaaleaacaWGHbaabeaakiaadk % gacqGH9aqpcaaIYaaabaGaeyi1HSTaamOyaiabg2da9iaadggadaah % aaWcbeqaaiaaikdaaaaakeaacqGHuhY2caWGHbGaeyypa0ZaaOaaae % aacaWGIbaaleqaaaaaaa!5730! \begin{array}{l} \,\,\,\,\,\,\frac{{{{\log }_2}b}}{{{{\log }_2}a}} = \frac{{2c}}{2}\\ \Leftrightarrow {\log _a}b = 2\\ \Leftrightarrow b = {a^2}\\ \Leftrightarrow a = \sqrt b \end{array}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaa % aaaaWdbiaadAgacaGGOaGaamiEaiaacMcacqGH9aqpdaWcaaWdaeaa % peGaaGymaaWdaeaapeGaaG4maaaacaWG4bWdamaaCaaaleqabaWdbi % aaiodaaaGccqGHRaWkcaWG4bWdamaaCaaaleqabaWdbiaaikdaaaGc % cqGHsislcaaIZaGaamiEaiabgUcaRiaaigdaaeaacaWGMbGaai4jai % aacIcacaWG4bGaaiykaiabg2da9iaadIhadaahaaWcbeqaaiaaikda % aaGccqGHRaWkcaaIYaGaamiEaiabgkHiTiaaiodacqGH9aqpcaaIWa % Gaeyi1HS9aamqaaqaabeqaaiaadIhadaWgaaWcbaGaaGymaaqabaGc % cqGH9aqpcqGHsislcaaIZaaabaGaamiEamaaBaaaleaacaaIYaaabe % aakiabg2da9iaaigdaaaGaay5waaaaaaa!5DFF! \begin{array}{l} f(x) = \frac{1}{3}{x^3} + {x^2} - 3x + 1\\ f'(x) = {x^2} + 2x - 3 = 0 \Leftrightarrow \left[ \begin{array}{l} {x_1} = - 3\\ {x_2} = 1 \end{array} \right. \end{array}\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaWG4bWaa0baaSqaaiaaigdaaeaacaaIZaaaaOGaey4kaSIaamiE % amaaDaaaleaacaaIYaaabaGaaG4maaaakiabg2da9iabgkHiTiaaik % dacaaI2aaaaa!3FBD! x_1^3 + x_2^3 = - 26\)

ĐK: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGeb % Gaeyypa0JaeSyhHeQaaiixaiaabUhacqGHsislcaaIXaGaai4oaiaa % iodacaGG9baabaWaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadI % hacqGHsgIRcqGHRaWkcqGHEisPaeqaaOGaamyEaiabg2da9iaaigda % caGG7aGaaGPaVpaaxababaGaciiBaiaacMgacaGGTbaaleaacaWG4b % GaeyOKH4QaeyOeI0IaeyOhIukabeaakiaadMhacqGH9aqpcaaIXaaa % aaa!57AF! \begin{array}{l} D =\mathbb{R} \backslash {\rm{\{ }} - 1;3\} \\ \end{array}\)

\(\mathrm{Ta\,có}\mathop {\lim }\limits_{x \to + \infty } y = 1;\,\mathop {\lim }\limits_{x \to - \infty } y = 1\) nên \(y=1\) là đường tiệm cận ngang của đồ thị hàm số.

Lại có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaGGOaGaeyOeI0Ia % aGymaiaacMcadaahaaadbeqaaiabgUcaRaaaaSqabaGccaWG5bGaey % ypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaGG7aGaaGPaVpaaxaba % baGaciiBaiaacMgacaGGTbaaleaacaWG4bGaeyOKH4Qaaiikaiabgk % HiTiaaigdacaGGPaWaaWbaaWqabeaacqGHsislaaaaleqaaOGaamyE % aiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaaaaa!5390! \mathop {\lim }\limits_{x \to {{( - 1)}^ + }} y = \frac{1}{2};\,\mathop {\lim }\limits_{x \to {{( - 1)}^ - }} y = \frac{1}{2}\) nên \(x=-1\) không phải là tiềm cận đứng của đồ thị hàm số

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaaIZaWaaWbaaWqa % beaacqGHRaWkaaaaleqaaOGaamyEaiabg2da9iabgUcaRiabg6HiLk % aacUdacaaMc8+aaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIha % cqGHsgIRcaaIZaWaaWbaaWqabeaacqGHsislaaaaleqaaOGaamyEai % abg2da9iabgkHiTiabg6HiLcaa!50AB! \mathop {\lim }\limits_{x \to {3^ + }} y = + \infty ;\,\mathop {\lim }\limits_{x \to {3^ - }} y = - \infty\) nên \(x=3\) là tiệm cận đứng của đồ thị hàm số.

Vậy đồ thị hàm số có 2 đường tiệm cận

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG6b % WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaadQhacqGHRaWk % caaIYaGaeyypa0JaaGimaiabgsDiBpaadeaaeaqabeaacaWG6bWaaS % baaSqaaiaaigdaaeqaaOGaeyypa0JaaGymaiabgUcaRiaadMgaaeaa % caWG6bWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaaGymaiabgkHiTi % aadMgaaaGaay5waaaabaGaamOEamaaDaaaleaacaaIXaaabaGaaGin % aaaakiabgUcaRiaadQhadaqhaaWcbaGaaGOmaaqaaiaaisdaaaGccq % GH9aqpcaGGOaGaaGymaiabgUcaRiaadMgacaGGPaWaaWbaaSqabeaa % caaI0aaaaOGaey4kaSIaaiikaiaaigdacqGHsislcaWGPbGaaiykam % aaCaaaleqabaGaaGinaaaakiabg2da9iabgkHiTiaaiIdaaaaa!60F9! \begin{array}{l} {z^2} - 2z + 2 = 0 \Leftrightarrow \left[ \begin{array}{l} {z_1} = 1 + i\\ {z_2} = 1 - i \end{array} \right.\\ \mathrm{Khi\,đó}\,z_1^4 + z_2^4 = {(1 + i)^4} + {(1 - i)^4} = - 8 \end{array}\)

Gọi \(O = AC \cap BD\)

Kẻ \(AH \bot SO \Rightarrow AH \bot \left( {SBD} \right) \Rightarrow d\left( {A,\left( {SBD} \right)} \right) = AH\)

Vì SA, AB, AD đôi một vuông góc nên ta có:

\(\frac{1}{{H{A^2}}} = \frac{1}{{S{A^2}}} + \frac{1}{{A{B^2}}} + \frac{1}{{A{D^2}}} = \frac{3}{{{a^2}}} \Rightarrow AH = \frac{{a\sqrt 3 }}{3}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG1bGaeyypa0ZaaOaaa8aabaWdbiaadIhaaSqabaGccqGHshI3 % caWG1bWaaWbaaSqabeaacaaIYaaaaOGaeyypa0JaamiEaiabgkDiEl % aaikdacaWG1bGaamizaiaadwhacqGH9aqpcaWGKbGaamiEaaaa!4883! % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaWG1bGaeyypa0ZaaOaaaeaacaWG4baaleqaaOGaeyO0H4Taamiz % aiaadwhacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaWaaOaaaeaaca % WG4baaleqaaaaakiaadsgacaWG4bGaeyO0H4TaaGOmaiaadsgacaWG % 1bGaeyypa0ZaaSaaaeaacaaIXaaabaWaaOaaaeaacaWG4baaleqaaa % aakiaadsgacaWG4baaaa!4CDF! u = \sqrt x \Rightarrow du = \frac{1}{{2\sqrt x }}dx \Rightarrow 2du = \frac{1}{{\sqrt x }}dx\)

Đổi cận

\(\begin{array}{c|c} x & 1 \,\,\,\,\,\,\,\,\,\,4 \\ \hline u & 1\,\,\,\,\,\,\,\,\,\,2 \end{array}\)

Khí đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGjbGaeyypa0Zaa8qCaeaacaWGLbWaaWbaaSqabeaacaWG1baa % aOGaaiOlaiaaikdacaWGKbGaamyDaiabg2da9aWcbaGaaGymaaqaai % aaikdaa0Gaey4kIipakiaaikdadaWdXbqaaiaadwgadaahaaWcbeqa % aiaadwhaaaGccaWGKbGaamyDaaWcbaGaaGymaaqaaiaaikdaa0Gaey % 4kIipaaaa!4ADD! % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWdXbqaamaalaaabaGaamyzamaaCaaaleqabaWaaOaaaeaacaWG % 4baameqaaaaaaOqaamaakaaabaGaamiEaaWcbeaaaaGccaWGKbGaam % iEaaWcbaGaaGymaaqaaiaaisdaa0Gaey4kIipaaaa!3F47! \int\limits_1^4 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} = \int\limits_1^2 {{e^u}.2du = } 2\int\limits_1^2 {{e^u}du} \)

TXĐ: \(D=\mathbb{R}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaa % aaaaWdbiaadMhacqGH9aqpcaWG4bWdamaaCaaaleqabaWdbiaaioda % aaGccqGHRaWkcaGGOaGaamyBaiabgUcaRiaaigdacaGGPaGaamiEa8 % aadaahaaWcbeqaa8qacaaIYaaaaOGaey4kaSIaaG4maiaadIhacqGH % RaWkcaaIYaaabaGaamyEaiaacEcacqGH9aqpcaaIZaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaaikdacaGGOaGaamyBaiabgUca % RiaaigdacaGGPaGaamiEaiabgUcaRiaaiodaaeaacqGHuoarcaGGNa % Gaeyypa0Jaaiikaiaad2gacqGHRaWkcaaIXaGaaiykamaaCaaaleqa % baGaaGOmaaaakiabgkHiTiaaiodacaGGUaGaaG4maiabg2da9iaad2 % gadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamyBaiabgkHi % TiaaiIdaaaaa!64F6! \begin{array}{l} y = {x^3} + (m + 1){x^2} + 3x + 2\\ y' = 3{x^2} + 2(m + 1)x + 3\\ \Delta = {4(m + 1)^2} - 4.3.3 = 4{m^2} + 8m - 32 \end{array}\)

Để hàm số đồng biến trên \(\mathbb{R}\) thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacqGHuoarcaGGNaGaeyizImQaaGimaiaaykW7caaMc8UaaGPaVlab % gcGiIiaadIhacqGHiiIZcqWIDesOaaa!43F6! y' % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyzImlaaa!37B9! \ge 0\,\,\,\forall x \in \mathbb{R}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuh % Y2daGabaabaeqabaGaamyyaiabg2da9iaaiodacqGH+aGpcaaIWaGa % aGPaVlaaykW7caGGOaGaamiBaiaadYgacaqGKbGaaiykaaqaaiabgs % 5aejabgsMiJkaaicdaaaGaay5EaaaaqaaaaaaaaaWdbeaacqGHuoar % cqGHKjYOcaaIWaGaeyi1HSTaamyBamaaCaaaleqabaGaaGOmaaaaki % abgUcaRiaaikdacaWGTbGaeyOeI0IaaGioaiabgsMiJkaaicdaaeaa % caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabgsDiBlabgkHi % TiaaisdacqGHKjYOcaWGTbGaeyizImQaaGOmaaaaaa!768F! \begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} a = 3 > 0\,\,(ll{\rm{d}})\\ \Delta \le 0 \end{array} \right.\\ \Delta \le 0 \Leftrightarrow 4{m^2} + 8m - 32 \le 0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow - 4 \le m \le 2 \end{array}\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGTbGaeyicI48aamWaaeaacqGHsislcaaI0aGaai4oaiaaikda % aiaawUfacaGLDbaaaaa!3DA1! m \in \left[ { - 4;2} \right]\)

Điều kiện \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG4bGaey4kaSIaaGOmaiaad2gacqGH+aGpcaaIWaaaaa!3B62! x + 2m > 0\)

Ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqababaaaaaaa % aapeqaaiaaikdadaahaaWcbeqaaiaadIhacqGHsislcaaIXaaaaOGa % eyypa0JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaisdaaeqaaOGaai % ikaiaadIhacqGHRaWkcaaIYaGaamyBaiaacMcacqGHRaWkcaWGTbaa % baGaeyi1HSTaaGOmamaaCaaaleqabaGaamiEaaaakiabg2da9iGacY % gacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakiaacIcacaWG4bGa % ey4kaSIaaGOmaiaad2gacaGGPaGaey4kaSIaaGOmaiaad2gaaaaa!55C0! \begin{array}{l} \,\,\,\,\,\,\,{2^{x - 1}} = {\log _4}(x + 2m) + m\\ \Leftrightarrow {2^x} = {\log _2}(x + 2m) + 2m \end{array}\)

đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaWG0bGaeyypa0JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikda % aeqaaOGaaiikaiaadIhacqGHRaWkcaaIYaGaamyBaiaacMcaaaa!40BA! t = {\log _2}(x + 2m)\)

khi đó có hệ phương trình: 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aadaGabaabaeqabaGaaGOmamaaCaaaleqabaGaamiEaaaakiabg2da % 9iaadshacqGHRaWkcaaIYaGaamyBaaqaaiaadshacqGH9aqpciGGSb % Gaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGccaGGOaGaamiEaiab % gUcaRiaaikdacaWGTbGaaiykaaaacaGL7baacqGHuhY2daGabaabae % qabaGaaGOmamaaCaaaleqabaGaamiEaaaakiabg2da9iaadshacqGH % RaWkcaaIYaGaamyBaaqaaiaaikdadaahaaWcbeqaaiaadshaaaGccq % GH9aqpcaWG4bGaey4kaSIaaGOmaiaad2gaaaGaay5EaaGaeyi1HS9a % aiqaaqaabeqaaiaaikdadaahaaWcbeqaaiaadIhaaaGccqGHsislca % WG0bGaeyypa0JaaGOmaiaad2gaaeaacaaIYaWaaWbaaSqabeaacaWG % 0baaaOGaeyOeI0IaamiEaiabg2da9iaaikdacaWGTbaaaiaawUhaaa % aa!6966! \left\{ \begin{array}{l} {2^x} = t + 2m\\ t = {\log _2}(x + 2m) \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {2^x} = t + 2m\\ {2^t} = x + 2m \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {2^x} - t = 2m\\ {2^t} - x = 2m \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacqGHuhY2caaIYaWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaamiD % aiabg2da9iaaikdadaahaaWcbeqaaiaadshaaaGccqGHsislcaWG4b % Gaeyi1HSTaaGPaVlaaikdadaahaaWcbeqaaiaadIhaaaGccqGHRaWk % caWG4bGaeyypa0JaaGOmamaaCaaaleqabaGaamiDaaaakiabgUcaRi % aadshacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caGGOaGa % aGymaiaacMcaaaa!58FA! \Leftrightarrow {2^x} - t = {2^t} - x \Leftrightarrow \,{2^x} + x = {2^t} + t\,\,\,\,\,\,(1)\)

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaWGMbGaaiikaiaadwhacaGGPaGaeyypa0JaaGOmamaaCaaaleqa % baGaamyDaaaakiabgUcaRiaadwhaaaa!3E20! f(u) = {2^u} + u\) là hàm số đồng biến trên \(\mathbb{R}\)

do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aadaqadaqaaiaaigdaaiaawIcacaGLPaaacqGHuhY2caWG4bGaeyyp % a0JaamiDaaaa!3DAF! \left( 1 \right) \Leftrightarrow x = t\)

vậy ta có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaaIYaWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaamiEaiabg2da % 9iaaikdacaWGTbaaaa!3CA1! {2^x} - x = 2m\)

Xét hàm số \(g(x)=2^x-x\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqababaaaaaaa % aapeqaaiaadEgacaGGNaGaaiikaiaadIhacaGGPaGaeyypa0JaaGOm % amaaCaaaleqabaGaamiEaaaakiGacYgacaGGUbGaaGOmaiabgkHiTi % aaigdaaeaacaWGNbGaai4jaiaacIcacaWG4bGaaiykaiabg2da9iaa % icdacqGHuhY2caaIYaWaaWbaaSqabeaacaWG4baaaOGaciiBaiaac6 % gacaaIYaGaeyOeI0IaaGymaiabg2da9iaaicdacqGHuhY2caWG4bGa % eyypa0JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaae % WaaeaadaWcaaqaaiaaigdaaeaaciGGSbGaaiOBaiaaikdaaaaacaGL % OaGaayzkaaGaeyypa0JaeyOeI0IaciiBaiaac+gacaGGNbWaaSbaaS % qaaiaaikdaaeqaaOWaaeWaaeaaciGGSbGaaiOBaiaaikdaaiaawIca % caGLPaaaaaaa!6839! \begin{array}{l} g'(x) = {2^x}\ln 2 - 1\\ g'(x) = 0 \Leftrightarrow {2^x}\ln 2 - 1 = 0 \Leftrightarrow x = {\log _2}\left( {\frac{1}{{\ln 2}}} \right) = - {\log _2}\left( {\ln 2} \right) \end{array}\)

bảng biến thiên của hàm số \(g(x)\)

nhìn vào bảng biến thiên ta thấy phương trình (1) có nghiệm

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuh % Y2caaIYaGaamyBaiabgwMiZoaalaaabaGaaGymaaqaaiGacYgacaGG % UbGaaGOmaaaacqGHRaWkciGGSbGaai4BaiaacEgadaWgaaWcbaGaaG % OmaaqabaGccaGGOaGaciiBaiaac6gacaaIYaGaaiykaaqaaiabgsDi % Blaad2gacqGHLjYSdaWcaaqaaiaaigdaaeaacaaIYaaaamaabmaaba % WaaSaaaeaacaaIXaaabaGaciiBaiaac6gacaaIYaaaaiabgUcaRiGa % cYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakiaacIcaciGGSb % GaaiOBaiaaikdacaGGPaaacaGLOaGaayzkaaGaeyisISRaaGimaiaa % cYcacaaI1aaaaaa!5FD8! \begin{array}{l} \Leftrightarrow 2m \ge \frac{1}{{\ln 2}} + {\log _2}(\ln 2)\\ \Leftrightarrow m \ge \frac{1}{2}\left( {\frac{1}{{\ln 2}} + {{\log }_2}(\ln 2)} \right) \approx 0,5 \end{array}\)

Kết hợp với điều kiện \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WGTbaacaGLhWUaayjcSdGaeyipaWJaaGymaiaaicdaaaa!3C80! \left| m \right| < 10\) thì có 9 giá trị m để phương trình có nghiệm

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG5b % Gaeyypa0JaamOzaiaacIcacaWG4bWaaWbaaSqabeaacaaIYaaaaOGa % ey4kaSIaaG4maiaadIhacaGGPaaabaGaamyEaiaacEcacqGH9aqpca % GGOaGaaGOmaiaadIhacqGHRaWkcaaIZaGaaiykaiaadAgacaGGNaGa % aiikaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIZaGaam % iEaiaacMcaaaaa!4DA9! \begin{array}{l} y = f({x^2} + 3x), \,\mathrm{ta\,có}\,y' = (2x + 3)f'({x^2} + 3x) \end{array}\)

cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIWaGaeyi1HS9aamqaaqaabeqaaiaaikdacaWG4bGa % ey4kaSIaaG4maiabg2da9iaaicdaaeaacaWGMbGaai4jaiaacIcaca % WG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4maiaadIhacaGG % PaGaeyypa0JaaGimaaaacaGLBbaacqGHuhY2daWabaabaeqabaGaam % iEaiabg2da9maalaaabaGaeyOeI0IaaG4maaqaaiaaikdaaaaabaGa % amOzaiaacEcacaGGOaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgU % caRiaaiodacaWG4bGaaiykaiabg2da9iaaicdaaaGaay5waaaaaa!5C1F! y' = 0 \Leftrightarrow \left[ \begin{array}{l} 2x + 3 = 0\\ f'({x^2} + 3x) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{{ - 3}}{2}\\ f'({x^2} + 3x) = 0 \end{array} \right.\)

Dựa vào bảng biến thiên của \(f'(x)\):

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEaiaacMcacqGH9aqpcaaIWaGaeyi1HS9aamqaaqaa % beqaaiaadIhacqGH9aqpcqGHsislcaaIYaaabaGaamiEaiabg2da9i % aaigdaaeaacaWG4bGaeyypa0JaaG4maaaacaGLBbaaaaa!4828! f'(x) = 0 \Leftrightarrow \left[ \begin{array}{l} x = - 2\\ x = 1\\ x = 3 \end{array} \right.\)

suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaioda % caWG4bGaaiykaiabg2da9iaaicdacqGHuhY2daWabaabaeqabaGaam % iEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaWG4bGaeyyp % a0JaeyOeI0IaaGOmaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkcaaIZaGaamiEaiabg2da9iaaigdaaeaacaWG4bWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSIaaG4maiaadIhacqGH9aqpcaaIZaaaai % aawUfaaiabgsDiBpaadeaaeaqabeaacaWG4bGaeyypa0JaeyOeI0Ia % aGymaaqaaiaadIhacqGH9aqpcqGHsislcaaIYaaabaGaamiEaiabg2 % da9maalaaabaGaeyOeI0IaaG4maiabgglaXoaakaaabaGaaGymaiaa % iodaaSqabaaakeaacaaIYaaaaaqaaiaadIhacqGH9aqpdaWcaaqaai % abgkHiTiaaiodacqGHXcqSdaGcaaqaaiaaikdacaaIXaaaleqaaaGc % baGaaGOmaaaaaaGaay5waaaaaa!7122! f'({x^2} + 3x) = 0 \Leftrightarrow \left[ \begin{array}{l} {x^2} + 3x = - 2\\ {x^2} + 3x = 1\\ {x^2} + 3x = 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = - 2\\ x = \frac{{ - 3 \pm \sqrt {13} }}{2}\\ x = \frac{{ - 3 \pm \sqrt {21} }}{2} \end{array} \right.\)

trong đó \(x = \frac{{ - 3 \pm \sqrt {13} }}{2}\) là nghiệm kép

từ bảng xét dấu của \(f'(x)\) ta có thể suy ra:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI % ZaGaamiEaaGaayjkaiaawMcaaiabgYda8iaaicdacqGHuhY2daWaba % abaeqabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaioda % caWG4bGaeyipaWJaeyOeI0IaaGOmaaqaaiaadIhadaahaaWcbeqaai % aaikdaaaGccqGHRaWkcaaIZaGaamiEaiabg6da+iaaiodaaaGaay5w % aaGaeyi1HS9aamqaaqaabeqaaiabgkHiTiaaikdacqGH8aapcaWG4b % GaeyipaWJaeyOeI0IaaGymaaqaaiaadIhacqGH8aapdaWcaaqaaiab % gkHiTiaaiodacqGHsisldaGcaaqaaiaaikdacaaIXaaaleqaaaGcba % GaaGOmaaaaaeaacaWG4bGaeyOpa4ZaaSaaaeaacqGHsislcaaIZaGa % ey4kaSYaaOaaaeaacaaIYaGaaGymaaWcbeaaaOqaaiaaikdaaaaaai % aawUfaaaaa!67F2! f'\left( {{x^2} + 3x} \right) < 0 \Leftrightarrow \left[ \begin{array}{l} {x^2} + 3x < - 2\\ {x^2} + 3x > 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2 < x < - 1\\ x < \frac{{ - 3 - \sqrt {21} }}{2}\\ x > \frac{{ - 3 + \sqrt {21} }}{2} \end{array} \right.\)

Ta có bảng xét dấu của \(y'\)

Quan sát bảng biến thiên ta thấy hàm số \(y=f(x^2+3x)\) có các điểm cực tiểu là 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcqGHsislcaaIYaaabaGaamiEaiabg2da9maa % laaabaGaeyOeI0IaaG4maiabgUcaRmaakaaabaGaaGOmaiaaigdaaS % qabaaakeaacaaIYaaaaaaacaGL7baaaaa!41B7! \left\{ \begin{array}{l} x = - 2\\ x = \frac{{ - 3 + \sqrt {21} }}{2} \end{array} \right.\)

Vậy hàm số có hai điểm cực tiểu

Gọi \(M', N', P', Q' \) lần lượt là trung điểm của  \(AB, BC, CD, AD\). H là giao điểm của QN và SO

Gọi O là tâm mặt đáy. Khi đó \(SO \bot (ABCD)\) và \(SO = \frac{1}{2}AC = \frac{{a\sqrt 2 }}{2}\)

Thể tích khối chóp \(S.ABCD\) là: \( {V_{S.ABCD}} = \frac{1}{3}SO.{S_{ABCD}} = \frac{{{a^3}\sqrt 2 }}{6}\)

Ta có:

\( {S_{MNPQ}} = 2{S_{NPQ}} = 2.\frac{4}{9}{S_{N'P'Q'}} = \frac{2}{9}4{S_{N'P'Q'}} = \frac{2}{9}{S_{ABCD}}\)

\(d(O;(MNPQ)) = OH = \frac{1}{3}SO\)

Vậy thể tích \(S.MNPQ\) là

\(\begin{array}{l} {V_{S.MNPQ}} = \frac{1}{3}OH.{S_{MNPQ}} = \frac{1}{3}\frac{1}{3}SO.\frac{2}{9}{S_{ABCD}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{27}}{V_{S.ABCD}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2{a^3}\sqrt 2 }}{{27.6}} = \frac{{{a^3}\sqrt 2 }}{{81}} \end{array}\)

Xét \(4^x-(m+1)2^x+2m-3=0\,\,\,\,(1)\)

Đặt \(t=2^x\,\,\,\,(t>0)\). Khi đó (1) trở thành:

\(t^2-(m+1)t+2m-3=0\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuh % Y2caWG0bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamiDaiabgkHi % TiaaiodacqGH9aqpcaWGTbGaaiikaiaadshacqGHsislcaaIYaGaai % ykaaqaaiabgsDiBlaad2gacqGH9aqpdaWcaaqaaiaadshadaahaaWc % beqaaiaaikdaaaGccqGHsislcaWG0bGaeyOeI0IaaG4maaqaaiaads % hacqGHsislcaaIYaaaaaaaaa!5047! \begin{array}{l} \Leftrightarrow {t^2} - t - 3 = m(t - 2)\\ \Leftrightarrow m = \frac{{{t^2} - t - 3}}{{t - 2}} \end{array}\)

Đặt \(f(t)=\frac{{{t^2} - t - 3}}{{t - 2}}\)

TXĐ: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maabmaabaGaaGimaiaacUdacqGHRaWkcqGHEisPaiaawIcacaGL % PaaacaGGCbGaae4EaiaaikdacaGG9baaaa!40B2! D = \left( {0; + \infty } \right)\backslash {\rm{\{ }}2\} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiDaiaacMcacqGH9aqpdaWcaaqaaiaadshadaahaaWc % beqaaiaaikdaaaGccqGHsislcaaI0aGaamiDaiabgUcaRiaaisdaae % aadaqadaqaaiaadshacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWba % aSqabeaacaaIYaaaaaaakiabg6da+iaaicdacaaMc8UaaGPaVlaayk % W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabgcGiIiaadshacqGH % iiIZcaWGebaaaa!586F! f'(t) = \frac{{{t^2} - 4t + 4}}{{{{\left( {t - 2} \right)}^2}}} > 0\,\,\,\,\,\,\,\,\forall t \in D\)

Bảng biến thiên

(1) có hai nghiệm phân biệt \(x_1, x_2\) trái dấu nghĩa là:

 \(x_1<0<x_2\)\( \Leftrightarrow {t_1} < 1 < {t_2}\)

\(\Leftrightarrow \)đường thẳng \(y=m\) cắt đường thẳng \(y=f(t)\) tại hai điểm nằm về hai phía của đường thẳng \(t=1\)

\(\Leftrightarrow 1,5<m<3\)

Vậy có 1 giá trị nguyên của m thỏa mãn điều kiện

Gọi I là hình chiếu vuông góc của B lên SE \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OqaiaadMeacqGHLkIxcaWGtbGaamyraaaa!3D38! \Rightarrow BI \bot SE\,\,\,\,(1)\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadkeacaWGdbGaeyyPI4Laam4uaiaadgeaaeaacaWGcbGaam4q % aiabgwQiEjaadgeacaWGcbaaaiaawUhaaiabgkDiElaadkeacaWGdb % GaeyyPI4LaaiikaiaadofacaWGbbGaamOqaiaacMcacqGHshI3caWG % cbGaam4qaiabgwQiEjaadkeacaWGjbaaaa!514C! \left\{ \begin{array}{l} BC \bot SA\\ BC \bot AB \end{array} \right. \Rightarrow BC \bot (SAB) \Rightarrow BC \bot BI\,\,\,\,\,(2)\)

Từ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaig % dacaGGPaGaaiilaiaacIcacaaIYaGaaiykaiabgkDiElaadsgacaGG % OaGaam4uaiaadweacaGGSaGaamOqaiaadoeacaGGPaGaeyypa0Jaam % OqaiaadMeaaaa!45E7! (1),(2) \Rightarrow d(SE,BC) = BI\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqaiaadM % eacqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaikdaaSqabaGccaGG % UaGaamyyaaqaaiaaikdadaGcaaqaaiaaikdacaWGHbWaaWbaaSqabe % aacaaIYaaaaOGaey4kaSYaaSaaaeaacaWGHbWaaWbaaSqabeaacaaI % YaaaaaGcbaGaaGinaaaaaSqabaaaaOGaeyypa0ZaaSaaaeaacaWGHb % WaaOaaaeaacaaIYaaaleqaaaGcbaGaaG4maaaaaaa!4696! BI = \frac{{a\sqrt 2 .a}}{{2\sqrt {2{a^2} + \frac{{{a^2}}}{4}} }} = \frac{{a\sqrt 2 }}{3}\)

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapehabaGaaiikaiaaisdacaWG4bGaey4kaSIaaGOmaiaacMca % ciGGSbGaaiOBaiaadIhacaWGKbGaamiEaaWcbaGaaGOmaaqaaiaaik % daa0Gaey4kIipaaaa!4521! I = \int\limits_2^2 {(4x + 2)\ln xdx} \)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpciGGSbGaaiOBaiaadIhaaeaacaWGKbGaamOD % aiabg2da9iaacIcacaaI0aGaamiEaiabgUcaRiaaikdacaGGPaGaam % izaiaadIhaaaGaay5EaaGaeyO0H49aaiqaaqaabeqaaiaadsgacaWG % 1bGaeyypa0ZaaSaaaeaacaaIXaaabaGaamiEaaaacaWGKbGaamiEaa % qaaiaadAhacqGH9aqpcaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaa % kiabgUcaRiaaikdacaWG4baaaiaawUhaaaaa!56D4! \left\{ \begin{array}{l} u = \ln x\\ dv = (4x + 2)dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{x}dx\\ v = 2{x^2} + 2x \end{array} \right.\)

Khi đó

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maaeiaabaWaamWaaeaadaqadaqaaiaaikdacaWG4bWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSIaaGOmaiaadIhaaiaawIcacaGLPaaaci % GGSbGaaiOBaiaadIhaaiaawUfacaGLDbaaaiaawIa7amaaDaaaleaa % caaIYaaabaGaaG4maaaakiabgkHiTmaapehabaWaaeWaaeaacaaIYa % GaamiEaiabgUcaRiaaikdaaiaawIcacaGLPaaaaSqaaiaaikdaaeaa % caaIZaaaniabgUIiYdaaaa!505F! I = \left. {\left[ {\left( {2{x^2} + 2x} \right)\ln x} \right]} \right|_2^3 - \int\limits_2^3 {\left( {2x + 2} \right)} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaaMb8 % UaaGzaVlaaygW7caaMb8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGOm % aiaaisdaciGGSbGaaiOBaiaaiodacqGHsislcaaIXaGaaGOmaiGacY % gacaGGUbGaaGOmaiabgkHiTmaaeiaabaWaaeWaaeaacaWG4bWaaWba % aSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadIhaaiaawIcacaGLPa % aaaiaawIa7amaaDaaaleaacaaIYaaabaGaaG4maaaaaOqaaiaaykW7 % caaMc8UaaGPaVlaaykW7cqGH9aqpcaaIYaGaaGinaiGacYgacaGGUb % GaaGinaiabgkHiTiaaigdacaaIYaGaciiBaiaac6gacaaIYaGaeyOe % I0IaaG4naaaaaa!672F! \begin{array}{l} \,\,\, = 24\ln 3 - 12\ln 2 - \left. {\left( {{x^2} + 2x} \right)} \right|_2^3\\ \,\,\,\, = 24\ln 4 - 12\ln 2 - 7 \end{array}\)

Vậy \(a=-7, b=-12, c=24\)

\(a=b+c=5\)