Xét các số phức z, w thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6baacaGLhWUaayjcSdGaeyypa0JaaGOmaiaacYcadaabdaqaaiaa % dMgacaWG3bGaeyOeI0IaaGOmaiabgUcaRiaaiwdacaWGPbaacaGLhW % UaayjcSdGaeyypa0JaaGymaaaa!478C! \left| z \right| = 2,\left| {iw - 2 + 5i} \right| = 1\). Giá trị nhỏ nhất của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0Iaae4DaiaadQhacqGH % sislcaaI0aaacaGLhWUaayjcSdaaaa!3F99! \left| {{z^2} - {\rm{w}}z - 4} \right|\) bằng:
A. 4
B.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaada
% GcaaqaaiaaikdacaaI5aaaleqaaOGaeyOeI0IaaG4maaGaayjkaiaa
% wMcaaaaa!3ACB!
2(\sqrt {29} - 3)\)
C. 8
D. \(2(\sqrt {29} - 5)\)
Lời giải của giáo viên
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 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa % aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0JaaGOmaiaac6ca % caaI0aGaeyypa0JaaGioaaaa!3ECF! {T_{\min }} = 2.4 = 8\)
CÂU HỎI CÙNG CHỦ ĐỀ
Cho số phức z = -2+ i . Trong hình bên điểm biểu diễn số phức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WG6baaaaaa!3704! \overline z \) là:
Trong không gian Oxyz, cho hai điểm A(-2;-1;3) và B( 0 ; 3 ;1) . Gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % aHXoqyaiaawIcacaGLPaaaaaa!391C! \left( \alpha \right)\) là mặt phẳng trung trực của AB. Một vecto pháp tuyến của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % aHXoqyaiaawIcacaGLPaaaaaa!391C! \left( \alpha \right)\) có tọa độ là:
Từ các chữ số 1; 2; 3;…; 9 lập được bao nhiêu số có 3 chữ số đôi một khác nhau
Cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maabmaabaGaamiEaiab % gkHiTiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaiodaaaGccq % GHsislcaaIZaGaamiEaiabgUcaRiaaiodaaaa!43D3! f\left( x \right) = {\left( {x - 1} \right)^3} - 3x + 3\). Đồ thị hình bên là của hàm số có công thức:
Biết rằng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiaadw % gadaahaaWcbeqaaiaadIhaaaaaaa!3905! x{e^x}\) là một nguyên hàm của hàm số f(-x) trên khoảng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % GHsislcqGHEisPcaGG7aGaey4kaSIaeyOhIukacaGLOaGaayzkaaaa % aa!3CED! \left( { - \infty ; + \infty } \right)\). Gọi F(x) là một nguyên hàm của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWGLbWaaWbaaSqabeaa % caWG4baaaaaa!3C24! f'\left( x \right){e^x}\) thỏa mãn F(0) = 1, giá trị của F(-1) bằng:
Cho hình hộp ABCD.A'B'C'D' có thể tích bằng V.Gọi M, N, P, Q, E, F lần lượt là tâm các hình bình hành ABCD,A'B'C'D', ABA'B', BCB'C',DAA'D'. Thể tích khối đa diện có các đỉnh M, P, Q, E, F, N bằng:
Cho hình chóp SABCD có đáy ABCD là hình vuông cạnh a , SA = a và SA \(\bot\) (ABCD). Thể tích khối chóp SABCD bằng:
Trong không gian Oxyz, một vecto chỉ phương của đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaai % OoamaalaaabaGaamiEaiabgkHiTiaaigdaaeaacaaIXaaaaiabg2da % 9maalaaabaGaamyEaiabgUcaRiaaiodaaeaacaaIYaaaaiabg2da9m % aalaaabaGaamOEaiabgkHiTiaaiodaaeaacqGHsislcaaI1aaaaaaa % !4562! \Delta :\frac{{x - 1}}{1} = \frac{{y + 3}}{2} = \frac{{z - 3}}{{ - 5}}\) có tọa độ là:
Trong không gian Oxyz, cho đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaWcaaqaaiaadIhacqGHsislcaaIZaaabaGaaGOmaaaacqGH9aqp % daWcaaqaaiaadMhacqGHsislcaaI0aaabaGaaGymaaaacqGH9aqpda % WcaaqaaiaadQhacqGHsislcaaIYaaabaGaaGymaaaaaaa!4401! d:\frac{{x - 3}}{2} = \frac{{y - 4}}{1} = \frac{{z - 2}}{1}\) và 2 điểm A( 6;3;-2); B(1;0;-1). Gọi \(\Delta\) là đường thẳng đi qua B, vuông góc với d và thỏa mãn khoảng cách từ A đến \(\Delta\) là nhỏ nhất. Một vectơ chỉ phương của có tọa độ:
Bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaisdaaeqaaOWaaeWaaeaacaWG4bWaaWba % aSqabeaacaaIYaaaaOGaeyOeI0IaaG4maiaadIhaaiaawIcacaGLPa % aacqGH+aGpciGGSbGaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGc % daqadaqaaiaaiMdacqGHsislcaWG4baacaGLOaGaayzkaaaaaa!48D8! {\log _4}\left( {{x^2} - 3x} \right) > {\log _2}\left( {9 - x} \right)\) có bao nhiêu nghiệm nguyên?
Biết \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakiaa % dIhacqGHRaWkciGGZbGaaiyAaiaac6gacaaMc8UaamiEaiGacogaca % GGVbGaai4CaiaadIhacqGHRaWkcaaIXaaabaGaci4yaiaac+gacaGG % ZbWaaWbaaSqabeaacaaI0aaaaOGaamiEaiabgUcaRiGacohacaGGPb % GaaiOBaiaaykW7caWG4bGaci4yaiaac+gacaGGZbWaaWbaaSqabeaa % caaIZaaaaOGaamiEaaaacaWGKbGaamiEaaWcbaWaaSaaaeaacqaHap % aCaeaacaaI0aaaaaqaamaalaaabaGaeqiWdahabaGaaG4maaaaa0Ga % ey4kIipakiabg2da9iaadggacqGHRaWkcaWGIbGaciiBaiaac6gaca % aIYaGaey4kaSIaam4yaiGacYgacaGGUbWaaeWaaeaacaaIXaGaey4k % aSYaaOaaaeaacaaIZaaaleqaaaGccaGLOaGaayzkaaaaaa!6DBA! \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\cos }^2}x + \sin \,x\cos x + 1}}{{{{\cos }^4}x + \sin \,x{{\cos }^3}x}}dx} = a + b\ln 2 + c\ln \left( {1 + \sqrt 3 } \right)\), với a, b, c là các số hữu tỉ. Giá trị của abc bằng:
Gọi (D) là hình phẳng giới hạn bởi các đường \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaaikdadaahaaWcbeqaaiaadIhaaaGccaGGSaGaamyEaiabg2da % 9iaaicdacaGGSaGaamiEaiabg2da9iaaicdaaaa!40C3! y = {2^x},y = 0,x = 0\) và x = 2. Thể tích V của khối tròn xoay tạo thành khi quay (D) quanh trục Ox được xác định bởi công thức:
Có bao nhiêu số nguyên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabgI % GiopaabmaabaGaeyOeI0IaaGOmaiaaicdacaaIXaGaaGyoaiaacUda % caaIYaGaaGimaiaaigdacaaI5aaacaGLOaGaayzkaaaaaa!417B! a \in \left( { - 2019;2019} \right)\) để phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaciiBaiaac6gadaqadaqaaiaadIhacqGHRaWkcaaI1aaa % caGLOaGaayzkaaaaaiabgUcaRmaalaaabaGaaGymaaqaaiaaiodada % ahaaWcbeqaaiaadIhaaaGccqGHsislcaaIXaaaaiabg2da9iaadIha % cqGHRaWkcaWGHbaaaa!45DB! \frac{1}{{\ln \left( {x + 5} \right)}} + \frac{1}{{{3^x} - 1}} = x + a\) có hai nghiệm phân biệt?
Cho hình chóp đều S.ABCD có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGH9aqpcaaIYaGaamyyaiaacYcacaWGtbGaamyqaiabg2da9iaa % dggadaGcaaqaaiaaiwdaaSqabaaaaa!3F3D! AB = 2a,SA = a\sqrt 5 \) . Góc giữa hai mặt phẳng (SAB) và (ABCD) bằng:
Gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBa % aaleaacaaIXaaabeaakiaacYcacaWG6bWaaSbaaSqaaiaaikdaaeqa % aaaa!3A7B! {z_1},{z_2}\) là các nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWG6bGaey4kaSIaaG4m % aiabg2da9iaaicdaaaa!3DED! {z^2} - 2z + 3 = 0\). Modul của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaDa % aaleaacaaIXaaabaGaaG4maaaakiaac6cacaWG6bWaa0baaSqaaiaa % ikdaaeaacaaI0aaaaaaa!3BFA! z_1^3.z_2^4\) bằng: