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Hỏi bàiBiết 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:
A.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca
% aI3aaabaGaaGOmaaaaaaa!3781!
\frac{7}{2}\)
B.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca
% aI1aGaeyOeI0Iaamyzaaqaaiaaikdaaaaaaa!3956!
\frac{{5 - e}}{2}\)
C.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca
% aI3aGaeyOeI0Iaamyzaaqaaiaaikdaaaaaaa!3958!
\frac{{7 - e}}{2}\)
D.
\(% MathType!MTEF!2!1!+-
% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca
% aI1aaabaGaaGOmaaaaaaa!377F!
\frac{5}{2}\)
Lời giải của giáo viên
HocOn247.com
Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiaadw % gadaahaaWcbeqaaiaadIhaaaaaaa!3905! x{e^x}\) là một nguyên hàm của hàm số f(-x) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaamyzamaaCaaaleqabaGaamiEaaaaaOGaayjkaiaawMcaaiaa % cEcacqGH9aqpcaWGMbWaaeWaaeaacqGHsislcaWG4baacaGLOaGaay % zkaaaaaa!40A7! \left( {x{e^x}} \right)' = f\left( { - x} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % OzamaabmaabaGaeyOeI0IaamiEaaGaayjkaiaawMcaaiabg2da9iaa % dwgadaahaaWcbeqaaiaadIhaaaGccqGHRaWkcaWG4bGaamyzamaaCa % aaleqabaGaamiEaaaakiabg2da9iaadwgadaahaaWcbeqaaiaadIha % aaGcdaqadaqaaiaaigdacqGHRaWkcaWG4baacaGLOaGaayzkaaaaaa!4B16! \Leftrightarrow f\left( { - x} \right) = {e^x} + x{e^x} = {e^x}\left( {1 + x} \right)\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OzamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadwgadaah % aaWcbeqaaiabgkHiTiaadIhaaaGcdaqadaqaaiaaigdacqGHsislca % WG4baacaGLOaGaayzkaaaaaa!4401! \Rightarrow f\left( x \right) = {e^{ - x}}\left( {1 - x} \right)\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcqGH % sislcaWGLbWaaWbaaSqabeaacqGHsislcaWG4baaaOWaaeWaaeaaca % aIXaGaeyOeI0IaamiEaaGaayjkaiaawMcaaiabgkHiTiaadwgadaah % aaWcbeqaaiabgkHiTiaadIhaaaGccqGH9aqpcqGHsislcaWGLbWaaW % baaSqabeaacqGHsislcaWG4baaaOWaaeWaaeaacaaIYaGaeyOeI0Ia % amiEaaGaayjkaiaawMcaaiabg2da9maabmaabaGaamiEaiabgkHiTi % aaikdaaiaawIcacaGLPaaacaWGLbWaaWbaaSqabeaacqGHsislcaWG % 4baaaaaa!5AF4! \Rightarrow f'\left( x \right) = - {e^{ - x}}\left( {1 - x} \right) - {e^{ - x}} = - {e^{ - x}}\left( {2 - x} \right) = \left( {x - 2} \right){e^{ - x}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3caWGMbGaai4jamaabmaabaGaamiEaaGaayjkaiaawMcaaiaadwga % daahaaWcbeqaaiaadIhaaaGccqGH9aqpdaqadaqaaiaadIhacqGHsi % slcaaIYaaacaGLOaGaayzkaaGaamyzamaaCaaaleqabaGaeyOeI0Ia % amiEaaaakiaac6cacaWGLbWaaWbaaSqabeaacaWG4baaaOGaeyypa0 % JaamiEaiabgkHiTiaaikdaaeaacqGHshI3caWGgbWaaeWaaeaacaWG % 4baacaGLOaGaayzkaaGaeyypa0Zaa8qaaeaacaWGMbWaaeWaaeaaca % WG4baacaGLOaGaayzkaaGaamizaiaadIhacqGH9aqpdaWdbaqaamaa % bmaabaGaamiEaiabgkHiTiaaikdaaiaawIcacaGLPaaacaWGKbGaam % iEaiabg2da9maalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaOqa % aiaaikdaaaGaeyOeI0IaaGOmaiaadIhacqGHRaWkcaWGdbaaleqabe % qdcqGHRiI8aaWcbeqab0Gaey4kIipaaOqaaiaadAeadaqadaqaaiaa % icdaaiaawIcacaGLPaaacqGH9aqpcaaIXaGaeyO0H4Taam4qaiabg2 % da9iaaigdacqGHshI3caWGgbWaaeWaaeaacaWG4baacaGLOaGaayzk % aaGaeyypa0ZaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGcba % GaaGOmaaaacqGHsislcaaIYaGaamiEaiabgUcaRiaaigdaaeaacqGH % shI3caWGgbWaaeWaaeaacqGHsislcaaIXaaacaGLOaGaayzkaaGaey % ypa0ZaaSaaaeaadaqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaa % daahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaiabgkHiTiaaikdada % qadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGHRaWkcaaIXaGa % eyypa0ZaaSaaaeaacaaI3aaabaGaaGOmaaaaaaaa!9949! \begin{array}{l} \Rightarrow f'\left( x \right){e^x} = \left( {x - 2} \right){e^{ - x}}.{e^x} = x - 2\\ \Rightarrow F\left( x \right) = \int {f\left( x \right)dx = \int {\left( {x - 2} \right)dx = \frac{{{x^2}}}{2} - 2x + C} } \\ F\left( 0 \right) = 1 \Rightarrow C = 1 \Rightarrow F\left( x \right) = \frac{{{x^2}}}{2} - 2x + 1\\ \Rightarrow F\left( { - 1} \right) = \frac{{{{\left( { - 1} \right)}^2}}}{2} - 2\left( { - 1} \right) + 1 = \frac{7}{2} \end{array}\)
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:
.png)
Từ các chữ số 1; 2; 3;…; 9 lập được bao nhiêu số có 3 chữ số đôi một khác nhau
Cho f(x) và g(x) là các hàm số liên tục bất kì trên đoạn [a;b]. Mệnh đề nào sau đây đúng ?
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:
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:
Cho hàm số f(x) có đạo hàm liên tục trên R thỏa mãn f(0) =3 và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabgUcaRiaadAgadaqadaqaaiaa % ikdacqGHsislcaWG4baacaGLOaGaayzkaaGaeyypa0JaamiEamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWG4bGaey4kaSIaaGOm % aiaaykW7caaMc8UaaGPaVlabgcGiIiaadIhacqGHiiIZcqWIDesOaa % a!4FFD! f\left( x \right) + f\left( {2 - x} \right) = {x^2} - 2x + 2\,\,\,\forall x \in \). Tích phân \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WG4bGaamOzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWG % KbGaamiEaaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIipaaaa!40D2! \int\limits_0^2 {xf'\left( x \right)dx} \) bằng:
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:
.png)
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:
Cho hình chóp SABCD có đáy ABCD là hình vuông cạnh a , SA = a và SA \(\bot\) (ABCD). Thể tích khối chóp SABCD bằng:
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?
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:
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:
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:
Có bao nhiêu số nguyên m để phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaaiodacqGH9aqpcaWGTbGaamyzamaaCaaaleqabaGaamiEaaaa % aaa!3C9C! x + 3 = m{e^x}\) có 2 nghiệm phân biệt?
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:
