Tìm tất cả các giá trị thực của tham số a ( a > 0) thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaadaqada % qaaiaaikdadaahaaWcbeqaaiaadggaaaGccqGHRaWkdaWcaaqaaiaa % igdaaeaacaaIYaWaaWbaaSqabeaacaWGHbaaaaaaaOGaayjkaiaawM % caamaaCaaaleqabaGaaGOmaiaaicdacaaIXaGaaG4naaaakiabgsMi % JoaabmaabaGaaGOmamaaCaaaleqabaGaaGOmaiaaicdacaaIXaGaaG % 4naaaakiabgUcaRmaalaaabaGaaGymaaqaaiaaikdadaahaaWcbeqa % aiaaikdacaaIWaGaaGymaiaaiEdaaaaaaaGccaGLOaGaayzkaaWaaW % baaSqabeaacaWGHbaaaaaa!4F2D! {\left( {{2^a} + \frac{1}{{{2^a}}}} \right)^{2017}} \le {\left( {{2^{2017}} + \frac{1}{{{2^{2017}}}}} \right)^a}\).
A.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB
% PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x
% c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr
% Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaaIWa
% GaeyipaWJaamyyaiabgsMiJkaaikdacaaIWaGaaGymaiaaiEdaaaa!3E9F!
0 < a \le 2017\)
B.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB
% PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x
% c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr
% Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaaIXa
% GaeyipaWJaamyyaiabgYda8iaaikdacaaIWaGaaGymaiaaiEdaaaa!3DEF!
1 < a < 2017\)
C.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB
% PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x
% c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr
% Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaWGHb
% GaeyyzImRaaGOmaiaaicdacaaIXaGaaG4naaaa!3CF2!
a \ge 2017\)
D.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB
% PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x
% c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr
% Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaaIWa
% GaeyipaWJaamyyaiabgYda8iaaigdaaaa!3BB7!
0 < a < 1\)
Lời giải của giáo viên
Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaadaqada % qaaiaaikdadaahaaWcbeqaaiaadggaaaGccqGHRaWkdaWcaaqaaiaa % igdaaeaacaaIYaWaaWbaaSqabeaacaWGHbaaaaaaaOGaayjkaiaawM % caamaaCaaaleqabaGaaGOmaiaaicdacaaIXaGaaG4naaaakiabgsMi % JoaabmaabaGaaGOmamaaCaaaleqabaGaaGOmaiaaicdacaaIXaGaaG % 4naaaakiabgUcaRmaalaaabaGaaGymaaqaaiaaikdadaahaaWcbeqa % aiaaikdacaaIWaGaaGymaiaaiEdaaaaaaaGccaGLOaGaayzkaaWaaW % baaSqabeaacaWGHbaaaaaa!4F2D! {\left( {{2^a} + \frac{1}{{{2^a}}}} \right)^{2017}} \le {\left( {{2^{2017}} + \frac{1}{{{2^{2017}}}}} \right)^a}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacqGHsh % I3caaIYaGaaGimaiaaigdacaaI3aGaaeiBaiaab+gacaqGNbWaaSba % aSqaaiaaikdaaeqaaOWaaeWaaeaacaaIYaWaaWbaaSqabeaacaWGHb % aaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmamaaCaaaleqabaGa % amyyaaaaaaaakiaawIcacaGLPaaacqGHKjYOcaWGHbGaaeiBaiaab+ % gacaqGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaaIYaWaaWba % aSqabeaacaaIYaGaaGimaiaaigdacaaI3aaaaOGaey4kaSYaaSaaae % aacaaIXaaabaGaaGOmamaaCaaaleqabaGaaGOmaiaaicdacaaIXaGa % aG4naaaaaaaakiaawIcacaGLPaaaaaa!58A0! \Rightarrow 2017{\rm{lo}}{{\rm{g}}_2}\left( {{2^a} + \frac{1}{{{2^a}}}} \right) \le a{\rm{lo}}{{\rm{g}}_2}\left( {{2^{2017}} + \frac{1}{{{2^{2017}}}}} \right)\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacqGHsh % I3daWcaaqaaiaabYgacaqGVbGaae4zamaaBaaaleaacaaIYaaabeaa % kmaabmaabaGaaGOmamaaCaaaleqabaGaamyyaaaakiabgUcaRmaala % aabaGaaGymaaqaaiaaikdadaahaaWcbeqaaiaadggaaaaaaaGccaGL % OaGaayzkaaaabaGaamyyaaaacqGHKjYOdaWcaaqaaiaabYgacaqGVb % Gaae4zamaaBaaaleaacaaIYaaabeaakmaabmaabaGaaGOmamaaCaaa % leqabaGaaGOmaiaaicdacaaIXaGaaG4naaaakiabgUcaRmaalaaaba % GaaGymaaqaaiaaikdadaahaaWcbeqaaiaaikdacaaIWaGaaGymaiaa % iEdaaaaaaaGccaGLOaGaayzkaaaabaGaaGOmaiaaicdacaaIXaGaaG % 4naaaaaaa!58C0! \Rightarrow \frac{{{\rm{lo}}{{\rm{g}}_2}\left( {{2^a} + \frac{1}{{{2^a}}}} \right)}}{a} \le \frac{{{\rm{lo}}{{\rm{g}}_2}\left( {{2^{2017}} + \frac{1}{{{2^{2017}}}}} \right)}}{{2017}}\)
Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaWG5b % Gaeyypa0JaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da % 9maalaaabaGaaeiBaiaab+gacaqGNbWaaSbaaSqaaiaaikdaaeqaaO % WaaeWaaeaacaaIYaWaaWbaaSqabeaacaWG4baaaOGaey4kaSYaaSaa % aeaacaaIXaaabaGaaGOmamaaCaaaleqabaGaamiEaaaaaaaakiaawI % cacaGLPaaaaeaacaWG4baaaiabg2da9maalaaabaGaaeiBaiaab+ga % caqGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaaI0aWaaWbaaS % qabeaacaWG4baaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaiabgkHi % TiaadIhaaeaacaWG4baaaiabg2da9maalaaabaGaaeiBaiaab+gaca % qGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaaI0aWaaWbaaSqa % beaacaWG4baaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaaqaaiaadI % haaaGaeyOeI0IaaGymaaaa!6311! y = f\left( x \right) = \frac{{{\rm{lo}}{{\rm{g}}_2}\left( {{2^x} + \frac{1}{{{2^x}}}} \right)}}{x} = \frac{{{\rm{lo}}{{\rm{g}}_2}\left( {{4^x} + 1} \right) - x}}{x} = \frac{{{\rm{lo}}{{\rm{g}}_2}\left( {{4^x} + 1} \right)}}{x} - 1\)
Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaaceWG5b % GbauaacqGH9aqpdaWcaaqaaiaaigdaaeaacaqGSbGaaeOBaiaaikda % aaWaamWaaeaadaWcaaqaamaalaaabaWaaeWaaeaacaaI0aWaaWbaaS % qabeaacaWG4baaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaiaacEca % aeaacaaI0aWaaWbaaSqabeaacaWG4baaaOGaey4kaSIaaGymaaaaca % GGUaGaamiEaiabgkHiTiaabYgacaqGUbWaaeWaaeaacaaI0aWaaWba % aSqabeaacaWG4baaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaaqaai % aadIhadaahaaWcbeqaaiaaikdaaaaaaaGccaGLBbGaayzxaaGaeyyp % a0ZaaSaaaeaacaaIXaaabaGaaeiBaiaab6gacaaIYaaaamaadmaaba % WaaSaaaeaacaaI0aWaaWbaaSqabeaacaWG4baaaOGaaiOlaiaabYga % caqGUbGaaGinaiaac6cacaWG4bGaeyOeI0YaaeWaaeaacaaI0aWaaW % baaSqabeaacaWG4baaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaiaa % bYgacaqGUbWaaeWaaeaacaaI0aWaaWbaaSqabeaacaWG4baaaOGaey % 4kaSIaaGymaaGaayjkaiaawMcaaaqaaiaadIhadaahaaWcbeqaaiaa % ikdaaaGcdaqadaqaaiaaisdadaahaaWcbeqaaiaadIhaaaGccqGHRa % WkcaaIXaaacaGLOaGaayzkaaaaaaGaay5waiaaw2faaiabgYda8iaa % icdaaaa!76D6! y' = \frac{1}{{{\rm{ln}}2}}\left[ {\frac{{\frac{{\left( {{4^x} + 1} \right)'}}{{{4^x} + 1}}.x - {\rm{ln}}\left( {{4^x} + 1} \right)}}{{{x^2}}}} \right] = \frac{1}{{{\rm{ln}}2}}\left[ {\frac{{{4^x}.{\rm{ln}}4.x - \left( {{4^x} + 1} \right){\rm{ln}}\left( {{4^x} + 1} \right)}}{{{x^2}\left( {{4^x} + 1} \right)}}} \right] < 0\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaaceWG5b % GbauaacqGH9aqpdaWcaaqaaiaaigdaaeaacaqGSbGaaeOBaiaaikda % aaWaamWaaeaadaWcaaqaaiaaisdadaahaaWcbeqaaiaadIhaaaGcca % GGUaGaaeiBaiaab6gacaaI0aWaaWbaaSqabeaacaWG4baaaOGaeyOe % I0YaaeWaaeaacaaI0aWaaWbaaSqabeaacaWG4baaaOGaey4kaSIaaG % ymaaGaayjkaiaawMcaaiaabYgacaqGUbWaaeWaaeaacaaI0aWaaWba % aSqabeaacaWG4baaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaaqaai % aadIhadaahaaWcbeqaaiaaikdaaaGcdaqadaqaaiaaisdadaahaaWc % beqaaiaadIhaaaGccqGHRaWkcaaIXaaacaGLOaGaayzkaaaaaaGaay % 5waiaaw2faaiabgYda8iaaicdaaaa!5B06! y' = \frac{1}{{{\rm{ln}}2}}\left[ {\frac{{{4^x}.{\rm{ln}}{4^x} - \left( {{4^x} + 1} \right){\rm{ln}}\left( {{4^x} + 1} \right)}}{{{x^2}\left( {{4^x} + 1} \right)}}} \right] < 0\) \(;% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacqGHai % IicaWG4bGaeyOpa4JaaGimaaaa!3AE3! \forall x > 0\)
Nên y = f(x) là hàm giảm trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabaqaceGabiGaaeaabaWaaeaaeaaakeaadaqada % qaaiaaicdacaGG7aGaey4kaSIaeyOhIukacaGLOaGaayzkaaaaaa!3CA7! \left( {0; + \infty } \right)\).
Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaWGMb % WaaeWaaeaacaWGHbaacaGLOaGaayzkaaGaeyizImQaamOzamaabmaa % baGaaGOmaiaaicdacaaIXaGaaG4naaGaayjkaiaawMcaaaaa!41C9! f\left( a \right) \le f\left( {2017} \right)\), ( a > 0) khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaaIWa % GaeyipaWJaamyyaiabgsMiJkaaikdacaaIWaGaaGymaiaaiEdaaaa!3E9F! 0 < a \le 2017\).
CÂU HỎI CÙNG CHỦ ĐỀ
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadggacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaamOy % aiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGJbGaamiEai % abgUcaRiaaigdaaaa!42EC! y = a{x^3} + b{x^2} + cx + 1\) có bảng biến thiên như sau:
Mệnh đề nào dưới đây đúng?
Cho hình chóp S.ABCD có đáy ABCD là hình vuông tâm O cạnh a, SO vuông góc với mặt phẳng (ABCD) và SO = a. Khoảng cách giữa SC và AB bằng
Công thức nào sau đây là đúng với cấp số cộng có số hạng đầu \(u_1\), công sai d, \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBaiabgw % MiZkaaikdacaGGUaaaaa!3A1A! n \ge 2.\) ?
Cho đồ thị hàm số y = f(x) có đồ thị như hình vẽ. Hàm số y = f(x) đồng biến trên khoảng nào dưới đây?
Cho hàm số y = f(x) có bảng biến thiên như sau:
Mệnh đề nào dưới đây đúng?
Cho hình chóp S.ABCD có đáy ABCD là hình vuông cạnh a, tam giác SAB đều và nằm trong mặt phẳng vuông góc với đáy. Tính thể tích khối cầu ngoại tiếp khối chóp SABCD.
Cho hình tứ diện OABC có đáy OBC là tam giác vuông tại O,OB =a ,OC= \(a\sqrt3\) . Cạnh OA vuông góc với mặt phẳng (OBC), \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadg % eacqGH9aqpcaWGHbWaaOaaaeaacaaIZaaaleqaaaaa!3A52! OA = a\sqrt 3 \) gọi M là trung điểm của BC . Tính theo a khoảng cách h giữa hai đường thẳng AB và OM.
Trong không gian với hệ tọa độ Oxyz, cho hai điểm A( -3;1; -4) và B(1; -1;2). Phương trình mặt cầu (S) nhận AB làm đường kính là
Nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqaM5cvLHfij5gC1rhimfMBNvxyNvga7TNm951EYG % xlX0xFTWLzYf2y7ftF7HtF9adatCvAUfeBSjuyZL2yd9gzLbvyNv2C % aerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLD % harqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr % 0xc9pk0xbba9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYR % Xxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaaba % aaaaaaaapeGaaGOma8aadaahaaWcbeqaa8qacaaIYaGaamiEaiabgk % HiTiaaigdaaaGccqGHsisldaWcaaWdaeaapeGaaGymaaWdaeaapeGa % aGioaaaacqGH9aqpcaaIWaaaaa!4F78! {2^{2x - 1}} - \frac{1}{8} = 0\) là
Trong không gian Oxyz, cho hình thoi ABCD với A(-1;2;1) ; B (2;3;2). Tâm I của hình thoi thuộc đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaWcaaqaaiaadIhacqGHRaWkcaaIXaaabaGaeyOeI0IaaGymaaaa % cqGH9aqpdaWcaaqaaiaadMhaaeaacqGHsislcaaIXaaaaiabg2da9m % aalaaabaGaamOEaiabgkHiTiaaikdaaeaacaaIXaaaaaaa!4421! d:\frac{{x + 1}}{{ - 1}} = \frac{y}{{ - 1}} = \frac{{z - 2}}{1}\). Tọa độ đỉnh D là
Cho A(1;-3;2) và mặt phẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaqadaqaaiaadcfaaiaawIcacaGLPaaacaGG6aGaaGOmaiaadIha % cqGHsislcaWG5bGaey4kaSIaaG4maiaadQhacqGHsislcaaIXaGaey % ypa0JaaGimaaaa!42DA! \left( P \right):2x - y + 3z - 1 = 0\) . Viết phương trình tham số đường thẳng d đi qua A, vuông góc với (P)
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEaiabgUcaRiaaikdaaeaacaWG4bGaey4kaSIa % aGymaaaaaaa!3D3D! y = \frac{{x + 2}}{{x + 1}}\) có đồ thị là (C). Gọi d là khoảng cách từ giao điểm 2 tiệm cận của (C) đến một tiếp tuyến bất kỳ của (C). Giá trị lớn nhất có thể đạt được là:
Trong không gian với hệ tọa độ Oxyz cho đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaqaaeaadaaakeaacaWGKbGaaiOoamaalaaabaGaamiEai % abgUcaRiaaiodaaeaacaaIYaaaaiabg2da9maalaaabaGaamyEaiab % gkHiTiaaigdaaeaacaaIXaaaaiabg2da9maalaaabaGaamOEaiabgk % HiTiaaigdaaeaacqGHsislcaaIZaaaaaaa!40A4! d:\frac{{x + 3}}{2} = \frac{{y - 1}}{1} = \frac{{z - 1}}{{ - 3}}\). Hình chiếu vuông góc của d trên mặt phẳng (Oyz) là một đường thẳng có vectơ chỉ phương là
Trong không gian với hệ tọa độ Oxyx , cho đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaWcaaqaaiaadIhacqGHsislcaaIXaaabaGaaGymaaaacqGH9aqp % daWcaaqaaiaadMhacqGHsislcaaIYaaabaGaaGymaaaacqGH9aqpda % WcaaqaaiaadQhacqGHsislcaaIXaaabaGaaGOmaaaaaaa!43FB! d:\frac{{x - 1}}{1} = \frac{{y - 2}}{1} = \frac{{z - 1}}{2}\),A(2;1;4) . Gọi H(a;b;c) là điểm thuộc d sao cho AH có độ dài nhỏ nhất. Tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9iaadggadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaWGIbWaaWba % aSqabeaacaaIZaaaaOGaey4kaSIaam4yamaaCaaaleqabaGaaG4maa % aaaaa!3F1D! T = {a^3} + {b^3} + {c^3}\).
Tính tích phân \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % qadaqaaiaaikdacaWGHbGaamiEaiabgUcaRiaadkgaaiaawIcacaGL % PaaacaqGKbGaamiEaaWcbaGaaGymaaqaaiaaikdaa0Gaey4kIipaaa % a!41A8! \int\limits_1^2 {\left( {2ax + b} \right){\rm{d}}x} \)