Trắc nghiệm Vi phân Toán Lớp 11

Câu 1:

Tìm vi phân của hàm số f(x) = tan2x – sin²(x + 1) tại điểm x = -1 ứng với Δx = -0,02 xấp xỉ bằng:

A. -0.233

B. -0,212

C. -0.312

D. -,0231
Câu 2:

Vi phân của hàm số f(x) = sin(3x – 2) + cos(x² + 1) tại điểm x = 0 ứng với Δx = 0,5 xấp xỉ bằng:

A. -0,24

B. -0,624

C. -0,364

D. Đáp án khác
Câu 3:

Vi phân của hàm số \(f\left( x \right) = \frac{{3x - 2}}{{2 + 5x}}\) tại điểm x = -1 ứng với Δx = 0,01 xấp xỉ bằng

A. 0,18

B. 0,018

C. 0,17

D. 0,017
Câu 4:

Tìm vi phân của hàm số \(f\left( x \right) = {\left( {{x^3} - \sqrt x } \right)^2}\) tại điểm x = 1 ứng với Δx = 0,5.

A. 1

B. - 1

C. 0

D. 2
Câu 5:

Tìm vi phân của hàm số y = (x³ – 2x²)².

A. dy = (6x5 – 20x4 + 16x)dx.

B. dy = (6x5 – 20x4 + 16x3)dx.

C. dy = (6x5 + 20x4 - 16x2)dx.

D. dy = (6x5 – 20x4 - 16x3)dx.
Câu 6:

Tìm vi phân của hàm số \(y = \frac{{2t - 3}}{{t + 4}}\)

A. \(dy= \frac{{11}}{{{{\left( {t + 4} \right)}^2}}}dt\)

B. \(dy= \frac{{11}}{{{{\left( {t - 4} \right)}^2}}}dt\)

C. \(dy= \frac{1}{{{{\left( {t + 4} \right)}^2}}}dt\)

D. \(dy= \frac{6}{{{{\left( {t - 4} \right)}^2}}}dt\)
Câu 7:

Tìm vi phân của hàm số \(y = {x^2} + 3\sqrt x + \frac{1}{x}\)

A. \(dy = \left( {2x + \frac{3}{{2\sqrt x }} - \frac{1}{x}} \right)dx\)

B. \(dy= \left( {2x + \frac{3}{{2\sqrt x }} - \frac{1}{{{x^2}}}} \right)dx\)

C. \(dy = \left( {2x + \frac{3}{{\sqrt x }} - \frac{1}{x}} \right)dx\)

D. \(dy= \left( {2x + \frac{3}{{\sqrt x }} - \frac{1}{{{x^2}}}} \right)dx\)
Câu 8:

Cho hàm số y = sin²x. Vi phân của hàm số là:

A. dy = -sin2xdx.

B. dy = sin2xdx.

C. dy = -sinxdx.

D. dy = 2cosxdx.
Câu 9:

Cho hàm số y = sinx – 3cosx. Vi phân của hàm số là:

A. dy = (-cosx + 3sinx)dx.

B. dy = (-cosx - 3sinx)dx.

C. dy = (cosx + 3sinx)dx.

D. dy = -(cosx + 3sinx)dx.
Câu 10:

Cho hàm số \(y = \frac{{{x^2} + x + 1}}{{x - 1}}\). Vi phân của hàm số là:

A. \(dy=- \frac{{{x^2} - 2x - 2}}{{{{\left( {x - 1} \right)}^2}}}dx

B. \(dy= \frac{{2x + 1}}{{{{\left( {x - 1} \right)}^2}}}dx\)

C. \(dy=- \frac{{2x + 1}}{{{{\left( {x - 1} \right)}^2}}}dx\)

D. \(dy= \frac{{{x^2} - 2x - 2}}{{{{\left( {x - 1} \right)}^2}}}dx
Câu 11:

Cho hàm số \(y = \frac{{x + 2}}{{x - 1}}\). Vi phân của hàm số là:

A. \(dy = \frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}\)

B. \(dy= \frac{3}{{{{\left( {x - 1} \right)}^2}}}dx\)

C. \(dy= - \frac{3}{{{{\left( {x - 1} \right)}^2}}}dx\)

D. \(dy = - \frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}\)
Câu 12:

Cho hàm số \(y = \frac{1}{{3{x^3}}}\). Vi phân của hàm số là:

A. \(dy= \frac{1}{4}dx\)

B. \(dy= \frac{1}{{{x^4}}}dx\)

C. \(dy= - \frac{1}{{{x^4}}}dx\)

D. dy = x4dx.
Câu 13:

Cho hàm số y = x³– 5x + 6 . Vi phân của hàm số là:

A. dy = (3x2 – 5)dx.

B. dy = -(3x2 – 5)dx.

C. dy = (3x2 + 5)dx.

D. dy = -(3x2 + 5)dx.
Câu 14:

Xét hàm số \(y = f\left( x \right) = \sqrt {2 + {{\cos }^2}2x} \). Chọn câu đúng:

A. \(df\left( x \right) = \frac{{ - \sin 4x}}{{2\sqrt {1 + {{\cos }^2}2x} }}dx\)

B. \(df\left( x \right) = \frac{{ - \sin 4x}}{{\sqrt {1 + {{\cos }^2}2x} }}dx\)

C. \(df\left( x \right) = \frac{{ \cos 2x}}{{\sqrt {1 + {{\cos }^2}2x} }}dx\)

D. \(df\left( x \right) = \frac{{ - \sin 2x}}{{2\sqrt {1 + {{\cos }^2}2x} }}dx\)
Câu 15:

Tìm vi phân của các hàm số y = tan2x

A. dy = (1 + tan22x)dx

B. dy = (1 - tan22x)dx

C. dy = 2(1 - tan22x)dx

D. dy = 2(1 + tan22x)dx
Câu 16:

Tìm vi phân của các hàm số y = sin2x + sin³x

A. dy = (cos2x + 3sin2xcosx)dx

B. dy = (2cos2x + 3sin2xcosx)dx

C. dy = (2cos2x + sin2xcosx)dx

D. dy = (cos2x + sin2xcosx)dx
Câu 17:

Tìm vi phân của các hàm số y = (3x + 1)¹⁰

A. dy = 10(3x + 1)9dx

B. dy = 30(3x + 1)10dx

C. dy = 9(3x + 1)10dx

D. dy = 30(3x + 1)9dx
Câu 18:

Cho hàm số y = x³– 9x² + 12x - 5. Vi phân của hàm số là:

A. dy = (3x2 – 18x + 12)dx.

B. dy = (-3x2 – 18x + 12)dx.

C. dy = -(3x2 – 18x + 12)dx.

D. dy = (-3x2 + 18x– 12)dx.
Câu 19:

Tìm vi phân của các hàm số \(y = \sqrt {3x + 2} \)

A. \(dy = \frac{3}{{\sqrt {3x + 2} }}dx\)

B. \(dy = \frac{1}{{\sqrt {3x + 2} }}dx\)

C. \(dy = \frac{1}{{2\sqrt {3x + 2} }}dx\)

D. \(dy = \frac{3}{{2\sqrt {3x + 2} }}dx\)
Câu 20:

Tìm vi phân của các hàm số y = x³ + 2x²

A. dy = (3x2 – 4x) dx

B. dy = (3x2 + x)dx

C. dy = (3x2 + 2x) dx

D. dy = (3x2 + 4x)dx
Câu 21:

Cho hàm số y = f(x) = (x – 1)². Biểu thức nào sau đây chỉ vi phân của hàm số f(x)?

A. dy = 2(x – 1)dx.

B. dy = (x – 1)2dx.

C. dy = 2(x – 1).

D. dy = 2(x – 1)dx.
Câu 22:

Một vật chuyển động theo quy luật \(s=\frac{-1}{2} t^{2}+20 t\) với t (giây) là khoảng thời gian tính từ khi vật bắt đầu chuyển động và s (mét) là quãng đường vật đi được trong thời gian đó. Hỏi vận tốc tức thời của vật tại thời điểm t = 8 giây bằng bao nhiêu?

A. \(40 \mathrm{m} / \mathrm{s}\)

B. \(152 \mathrm{m} / \mathrm{s}\)

C. \(22 \mathrm{m} / \mathrm{s}\)

D. \(12 \mathrm{m} / \mathrm{s}\)
Câu 23:

Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG5bGaaeiiaiabg2da98aadaWcaaqaaiaadIhaaeaacaWG4bWa % aWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaaaaaaa!3D64! y{\rm{ }} = \frac{x}{{{x^2} + 1}}\). Có vi phân là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdacqGHsislcaWG4bWaaWbaaSqabeaa % caaIYaaaaaGcbaGaaiikaiaadIhadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkcaaIXaGaaiykamaaCaaaleqabaGaaGOmaaaaaaGccaWGKbGa % amiEaaaa!4447! dy = \frac{{1 - {x^2}}}{{{{({x^2} + 1)}^2}}}dx\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaikdacaWG4baabaGaaiikaiaadIhadaah % aaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaGaaiykaaaacaWGKbGaam % iEaaaa!4175! dy = \frac{{2x}}{{({x^2} + 1)}}dx\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdacqGHsislcaWG4bWaaWbaaSqabeaa % caaIYaaaaaGcbaGaaiikaiaadIhadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkcaaIXaGaaiykaaaacaWGKbGaamiEaaaa!4354! dy = \frac{{1 - {x^2}}}{{({x^2} + 1)}}dx\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdaaeaacaGGOaGaamiEamaaCaaaleqa % baGaaGOmaaaakiabgUcaRiaaigdacaGGPaWaaWbaaSqabeaacaaIYa % aaaaaakiaadsgacaWG4baaaa!416A! dy = \frac{1}{{{{({x^2} + 1)}^2}}}dx\)
Câu 24:

Hàm số y = x sinx + cosx có vi phân là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGKbGaamyEaiabg2da98aadaqadaqaa8qacaWG4bGaci4yaiaa % c+gacaGGZbGaamiEaiaacobiciGGZbGaaiyAaiaac6gacaWG4baapa % GaayjkaiaawMcaa8qacaqGKbGaamiEaaaa!4602! {\rm{d}}y = \left( {x\cos x--\sin x} \right){\rm{d}}x\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGKbGaamyEaiabg2da98aadaqadaqaa8qaciGGJbGaai4Baiaa % cohacaWG4bGaai4eGiGacohacaGGPbGaaiOBaiaadIhaa8aacaGLOa % GaayzkaaWdbiaabsgacaWG4baaaa!4505! {\rm{d}}y = \left( {\cos x--\sin x} \right){\rm{d}}x\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGKbGaamyEaiabg2da98aadaqadaqaa8qacaWG4bGaci4yaiaa % c+gacaGGZbGaamiEaaWdaiaawIcacaGLPaaapeGaaeizaiaadIhaaa % a!4176! {\rm{d}}y = \left( {x\cos x} \right){\rm{d}}x\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGKbGaamyEaiabg2da98aadaqadaqaa8qacaWG4bGaci4yaiaa % c+gacaGGZbGaamiEaaWdaiaawIcacaGLPaaapeGaaeizaiaadIhaaa % a!4176! {\rm{d}}y = \left( {x\sin x} \right){\rm{d}}x\)
Câu 25:

Vi phân của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaciiDaiaacggacaGGUbWaaOaaaeaacaWG4baaleqa % aaGcbaWaaOaaaeaacaWG4baaleqaaaaaaaa!3D12! y = \frac{{\tan \sqrt x }}{{\sqrt x }}\) là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiaaikdadaGcaaqaaiaadIhaaSqabaaakeaa % caaI0aGaamiEamaakaaabaGaamiEaaWcbeaakiGacogacaGGVbGaai % 4CamaaCaaaleqabaGaaGOmaaaakmaakaaabaGaamiEaaWcbeaaaaGc % caqGKbGaamiEaaaa!4475! {\rm{d}}y = \frac{{2\sqrt x }}{{4x\sqrt x {{\cos }^2}\sqrt x }}{\rm{d}}x\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiGacohacaGGPbGaaiOBaiaacIcacaaIYaWa % aOaaaeaacaWG4baaleqaaOGaaiykaaqaaiaaisdacaWG4bWaaOaaae % aacaWG4baaleqaaOGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaI % YaaaaOWaaOaaaeaacaWG4baaleqaaaaakiaabsgacaWG4baaaa!48A6! {\rm{d}}y = \frac{{\sin (2\sqrt x )}}{{4x\sqrt x {{\cos }^2}\sqrt x }}{\rm{d}}x\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiaaikdadaGcaaqaaiaadIhaaSqabaGccqGH % sislciGGZbGaaiyAaiaac6gacaGGOaGaaGOmamaakaaabaGaamiEaa % WcbeaakiaacMcaaeaacaaI0aGaamiEamaakaaabaGaamiEaaWcbeaa % kiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakmaakaaaba % GaamiEaaWcbeaaaaGccaqGKbGaamiEaaaa!4B71! {\rm{d}}y = \frac{{2\sqrt x - \sin (2\sqrt x )}}{{4x\sqrt x {{\cos }^2}\sqrt x }}{\rm{d}}x\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcqGHsisldaWcaaqaaiaaikdadaGcaaqaaiaadIhaaSqa % baGccqGHsislciGGZbGaaiyAaiaac6gacaGGOaGaaGOmamaakaaaba % GaamiEaaWcbeaakiaacMcaaeaacaaI0aGaamiEamaakaaabaGaamiE % aaWcbeaakiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakm % aakaaabaGaamiEaaWcbeaaaaGccaqGKbGaamiEaaaa!4C5E! {\rm{d}}y = - \frac{{2\sqrt x - \sin (2\sqrt x )}}{{4x\sqrt x {{\cos }^2}\sqrt x }}{\rm{d}}x\)
Câu 26:

Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG5bGaeyypa0Jaci4CaiaacMgacaGGUbWdamaaCaaaleqabaWd % biaaikdaaaGccaWG4baaaa!3CFE! y = {\sin ^2}x\). Vi phân của hàm số là:

A. dy = - sin2x dx

B. dy = sin2x dx

C. dy = sinx dx

D. dy = 2cosx dx
Câu 27:

Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBaiaadIhacqGHsislcaaIZaGaci4yaiaa % c+gacaGGZbGaamiEaaaa!4147! y = \sin x - 3\cos x\). Vi phân của hàm số là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaqadaqaaiabgkHiTiGacogacaGGVbGaai4CaiaadIha % cqGHRaWkcaaIZaGaci4CaiaacMgacaGGUbGaamiEaaGaayjkaiaawM % caaiaabsgacaWG4baaaa!467D! {\rm{d}}y = \left( { - \cos x + 3\sin x} \right){\rm{d}}x\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaqadaqaaiabgkHiTiGacogacaGGVbGaai4CaiaadIha % cqGHsislcaaIZaGaci4CaiaacMgacaGGUbGaamiEaaGaayjkaiaawM % caaiaabsgacaWG4baaaa!4688! {\rm{d}}y = \left( { - \cos x - 3\sin x} \right){\rm{d}}x\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaqadaqaaiGacogacaGGVbGaai4CaiaadIhacqGHRaWk % caaIZaGaci4CaiaacMgacaGGUbGaamiEaaGaayjkaiaawMcaaiaabs % gacaWG4baaaa!4590! {\rm{d}}y = \left( {\cos x + 3\sin x} \right){\rm{d}}x\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcqGHsisldaqadaqaaiGacogacaGGVbGaai4CaiaadIha % cqGHRaWkcaaIZaGaci4CaiaacMgacaGGUbGaamiEaaGaayjkaiaawM % caaiaabsgacaWG4baaaa!467D! {\rm{d}}y = - \left( {\cos x + 3\sin x} \right){\rm{d}}x\)
Câu 28:

Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaa % dIhacqGHRaWkcaaIXaaabaGaamiEaiabgkHiTiaaigdaaaaaaa!4019! y = \frac{{{x^2} + x + 1}}{{x - 1}}\). Vi phân của hàm số là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikda % aaGccqGHsislcaaIYaGaamiEaiabgkHiTiaaikdaaeaacaGGOaGaam % iEaiabgkHiTiaaigdacaGGPaWaaWbaaSqabeaacaaIYaaaaaaakiaa % bsgacaWG4baaaa!46F0! {\rm{d}}y = - \frac{{{x^2} - 2x - 2}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiaaikdacaWG4bGaey4kaSIaaGymaaqaaiaa % cIcacaWG4bGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaaiaaikdaaa % aaaOGaaeizaiaadIhaaaa!431A! {\rm{d}}y = \frac{{2x + 1}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcqGHsisldaWcaaqaaiaaikdacaWG4bGaey4kaSIaaGym % aaqaaiaacIcacaWG4bGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaai % aaikdaaaaaaOGaaeizaiaadIhaaaa!4407! {\rm{d}}y = - \frac{{2x + 1}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGH % sislcaaIYaGaamiEaiabgkHiTiaaikdaaeaacaGGOaGaamiEaiabgk % HiTiaaigdacaGGPaWaaWbaaSqabeaacaaIYaaaaaaakiaabsgacaWG % 4baaaa!4603! {\rm{d}}y = \frac{{{x^2} - 2x - 2}}{{{{(x - 1)}^2}}}{\rm{d}}x\)
Câu 29:

Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEaiabgUcaRiaaikdaaeaacaWG4bGaeyOeI0Ia % aGymaaaaaaa!3D48! y = \frac{{x + 2}}{{x - 1}}\). Vi phân của hàm số là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiaabsgacaWG4baabaWaaeWaaeaacaWG4bGa % eyOeI0IaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaa % aaaa!3FEA! {\rm{d}}y = \frac{{{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiaaiodacaqGKbGaamiEaaqaamaabmaabaGa % amiEaiabgkHiTiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaik % daaaaaaaaa!40A7! {\rm{d}}y = \frac{{3{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiabgkHiTiaaiodacaqGKbGaamiEaaqaamaa % bmaabaGaamiEaiabgkHiTiaaigdaaiaawIcacaGLPaaadaahaaWcbe % qaaiaaikdaaaaaaaaa!4194! {\rm{d}}y = \frac{{ - 3{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcqGHsisldaWcaaqaaiaabsgacaWG4baabaWaaeWaaeaa % caWG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaG % Omaaaaaaaaaa!40D7! {\rm{d}}y = - \frac{{{\rm{d}}x}}{{{{\left( {x - 1} \right)}^2}}}\)
Câu 30:

Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaaqaaiaaiodacaWG4bWaaWbaaSqabeaacaaI % Zaaaaaaaaaa!3B67! y = \frac{1}{{3{x^3}}}\). Vi phân của hàm số là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdaaeaacaaI0aaaaiaabsgacaWG4baa % aa!3C4C! {\rm{d}}y = \frac{1}{4}{\rm{d}}x\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdaaeaacaWG4bWaaWbaaSqabeaacaaI % 0aaaaaaakiaabsgacaWG4baaaa!3D80! {\rm{d}}y = \frac{1}{{{x^4}}}{\rm{d}}x\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaWG4bWaaWbaaSqa % beaacaaI0aaaaaaakiaabsgacaWG4baaaa!3E6D! {\rm{d}}y = - \frac{1}{{{x^4}}}{\rm{d}}x\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcaWG4bWaaWbaaSqabeaacaaI0aaaaOGaaeizaiaadIha % aaa!3CB5! {\rm{d}}y = {x^4}{\rm{d}}x\)
Câu 31:

Cho hàm số \( y = {x^3} - 5x + 6\) . Vi phân của hàm số là:

A. \({\rm{d}}y = \left( {3{x^2} - 5} \right){\rm{d}}x\)

B. \({\rm{d}}y = - \left( {3{x^2} - 5} \right){\rm{d}}x\)

C. \( {\rm{d}}y = \left( {3{x^2} + 5} \right){\rm{d}}x\)

D. \( {\rm{d}}y = -\left( {3{x^2} + 5} \right){\rm{d}}x\)
Câu 32:

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaGc % aaqaaiaaigdacqGHRaWkciGGJbGaai4BaiaacohadaahaaWcbeqaai % aaikdaaaGccaaIYaGaamiEaaWcbeaaaaa!43A6! y = f\left( x \right) = \sqrt {1 + {{\cos }^2}2x} \). Chọn câu đúng:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadA % gacaGGOaGaamiEaiaacMcacqGH9aqpdaWcaaqaaiabgkHiTiGacoha % caGGPbGaaiOBaiaaisdacaWG4baabaGaaGOmamaakaaabaGaaGymai % abgUcaRiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakiaa % ikdacaWG4baaleqaaaaakiaabsgacaWG4baaaa!4A93! {\rm{d}}f(x) = \frac{{ - \sin 4x}}{{2\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadA % gacaGGOaGaamiEaiaacMcacqGH9aqpdaWcaaqaaiabgkHiTiGacoha % caGGPbGaaiOBaiaaisdacaWG4baabaWaaOaaaeaacaaIXaGaey4kaS % Iaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaaGOmaiaa % dIhaaSqabaaaaOGaaeizaiaadIhaaaa!49D7! {\rm{d}}f(x) = \frac{{ - \sin 4x}}{{\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadA % gacaGGOaGaamiEaiaacMcacqGH9aqpdaWcaaqaaiGacogacaGGVbGa % ai4CaiaaikdacaWG4baabaWaaOaaaeaacaaIXaGaey4kaSIaci4yai % aac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaaGOmaiaadIhaaSqa % baaaaOGaaeizaiaadIhaaaa!48E3! {\rm{d}}f(x) = \frac{{\cos 2x}}{{\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadA % gacaGGOaGaamiEaiaacMcacqGH9aqpdaWcaaqaaiabgkHiTiGacoha % caGGPbGaaiOBaiaaikdacaWG4baabaGaaGOmamaakaaabaGaaGymai % abgUcaRiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakiaa % ikdacaWG4baaleqaaaaakiaabsgacaWG4baaaa!4A91! {\rm{d}}f(x) = \frac{{ - \sin 2x}}{{2\sqrt {1 + {{\cos }^2}2x} }}{\rm{d}}x\)
Câu 33:

Tìm vi phân của các hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maakeaabaGaamiEaiabgUcaRiaaigdaaSqaaiaaiodaaaaaaa!3B6A! y = \sqrt[3]{{x + 1}}\)

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcbaqaaiaacIcacaWG4bGa % ey4kaSIaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaaabaGaaG4maa % aaaaGccaWGKbGaamiEaaaa!4145! dy = \frac{1}{{\sqrt[3]{{{{(x + 1)}^2}}}}}dx\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaiodaaeaadaGcbaqaaiaacIcacaWG4bGa % ey4kaSIaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaaabaGaaG4maa % aaaaGccaWGKbGaamiEaaaa!4147! dy = \frac{3}{{\sqrt[3]{{{{(x + 1)}^2}}}}}dx\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaikdaaeaadaGcbaqaaiaacIcacaWG4bGa % ey4kaSIaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaaabaGaaG4maa % aaaaGccaWGKbGaamiEaaaa!4146! dy = \frac{2}{{\sqrt[3]{{{{(x + 1)}^2}}}}}dx\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIZaWaaOqaaeaacaGGOaGa % amiEaiabgUcaRiaaigdacaGGPaWaaWbaaSqabeaacaaIYaaaaaqaai % aaiodaaaaaaOGaamizaiaadIhaaaa!4202! dy = \frac{1}{{3\sqrt[3]{{{{(x + 1)}^2}}}}}dx\)
Câu 34:

Tìm vi phân của các hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacshacaGGHbGaaiOBaiaaikdacaWG4baaaa!3C82! y = \tan 2x\)

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaGGOaGaaGymaiabgUcaRiGacshacaGGHbGaaiOBamaa % CaaaleqabaGaaGOmaaaakiaaikdacaWG4bGaaiykaiaadsgacaWG4b % aaaa!433A! dy = (1 + {\tan ^2}2x)dx\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaGGOaGaaGymaiabgkHiTiGacshacaGGHbGaaiOBamaa % CaaaleqabaGaaGOmaaaakiaaikdacaWG4bGaaiykaiaadsgacaWG4b % aaaa!4345! dy = (1 - {\tan ^2}2x)dx\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaaIYaGaaiikaiaaigdacqGHsislciGG0bGaaiyyaiaa % c6gadaahaaWcbeqaaiaaikdaaaGccaaIYaGaamiEaiaacMcacaWGKb % GaamiEaaaa!4401! dy = 2(1 - {\tan ^2}2x)dx\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaaIYaGaaiikaiaaigdacqGHRaWkciGG0bGaaiyyaiaa % c6gadaahaaWcbeqaaiaaikdaaaGccaaIYaGaamiEaiaacMcacaWGKb % GaamiEaaaa!43F6! dy = 2(1 + {\tan ^2}2x)dx\)
Câu 35:

Tìm vi phân của các hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBaiaaikdacaWG4bGaey4kaSIaci4Caiaa % cMgacaGGUbWaaWbaaSqabeaacaaIZaaaaOGaamiEaaaa!4234! y = \sin 2x + {\sin ^3}x\)

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaqadaqaaiGacogacaGGVbGaai4CaiaaikdacaWG4bGa % ey4kaSIaaG4maiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaa % aakiaadIhaciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaGa % amizaiaadIhaaaa!4B13! dy = \left( {\cos 2x + 3{{\sin }^2}x\cos x} \right)dx\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaqadaqaaiaaikdaciGGJbGaai4BaiaacohacaaIYaGa % amiEaiabgUcaRiaaiodaciGGZbGaaiyAaiaac6gadaahaaWcbeqaai % aaikdaaaGccaWG4bGaci4yaiaac+gacaGGZbGaamiEaaGaayjkaiaa % wMcaaiaadsgacaWG4baaaa!4BCF! dy = \left( {2\cos 2x + 3{{\sin }^2}x\cos x} \right)dx\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaqadaqaaiaaikdaciGGJbGaai4BaiaacohacaaIYaGa % amiEaiabgUcaRiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaa % aakiaadIhaciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaGa % amizaiaadIhaaaa!4B12! dy = \left( {2\cos 2x + {{\sin }^2}x\cos x} \right)dx\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaqadaqaaiGacogacaGGVbGaai4CaiaaikdacaWG4bGa % ey4kaSIaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOGaam % iEaiGacogacaGGVbGaai4CaiaadIhaaiaawIcacaGLPaaacaWGKbGa % amiEaaaa!4A56! dy = \left( {\cos 2x + {{\sin }^2}x\cos x} \right)dx\)
Câu 36:

Tìm vi phân của các hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaacIcacaaIZaGaamiEaiabgUcaRiaaigdacaGGPaWaaWbaaSqa % beaacaaIXaGaaGimaaaaaaa!3E4A! y = {(3x + 1)^{10}}\)

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaaIXaGaaGimaiaacIcacaaIZaGaamiEaiabgUcaRiaa % igdacaGGPaWaaWbaaSqabeaacaaI5aaaaOGaamizaiaadIhaaaa!41E6! dy = 10{(3x + 1)^9}dx\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaaIZaGaaGimaiaacIcacaaIZaGaamiEaiabgUcaRiaa % igdacaGGPaWaaWbaaSqabeaacaaIXaGaaGimaaaakiaadsgacaWG4b % aaaa!429A! dy = 30{(3x + 1)^{10}}dx\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaaI5aGaaiikaiaaiodacaWG4bGaey4kaSIaaGymaiaa % cMcadaahaaWcbeqaaiaaigdacaaIWaaaaOGaamizaiaadIhaaaa!41E6! dy = 9{(3x + 1)^{10}}dx\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaaIZaGaaGimaiaacIcacaaIZaGaamiEaiabgUcaRiaa % igdacaGGPaWaaWbaaSqabeaacaaI5aaaaOGaamizaiaadIhaaaa!41E8! dy = 30{(3x + 1)^9}dx\)
Câu 37:

Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaI5aGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigdacaaIYaGaamiEai % abgkHiTiaaiwdaaaa!428B! y = {x^3} - 9{x^2} + 12x - 5\). Vi phân của hàm số là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaqadaqaaiaaiodacaWG4bWaaWbaaSqabeaacaaIYaaa % aOGaeyOeI0IaaGymaiaaiIdacaWG4bGaey4kaSIaaGymaiaaikdaai % aawIcacaGLPaaacaqGKbGaamiEaaaa!44B9! {\rm{d}}y = \left( {3{x^2} - 18x + 12} \right){\rm{d}}x\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaqadaqaaiabgkHiTiaaiodacaWG4bWaaWbaaSqabeaa % caaIYaaaaOGaeyOeI0IaaGymaiaaiIdacaWG4bGaey4kaSIaaGymai % aaikdaaiaawIcacaGLPaaacaqGKbGaamiEaaaa!45A6! {\rm{d}}y = \left( { - 3{x^2} - 18x + 12} \right){\rm{d}}x\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcqGHsisldaqadaqaaiaaiodacaWG4bWaaWbaaSqabeaa % caaIYaaaaOGaeyOeI0IaaGymaiaaiIdacaWG4bGaey4kaSIaaGymai % aaikdaaiaawIcacaGLPaaacaqGKbGaamiEaaaa!45A6! {\rm{d}}y = - \left( {3{x^2} - 18x + 12} \right){\rm{d}}x\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaqadaqaaiabgkHiTiaaiodacaWG4bWaaWbaaSqabeaa % caaIYaaaaOGaey4kaSIaaGymaiaaiIdacaWG4bGaeyOeI0IaaGymai % aaikdaaiaawIcacaGLPaaacaqGKbGaamiEaaaa!45A6! {\rm{d}}y = \left( { - 3{x^2} + 18x - 12} \right){\rm{d}}x\)
Câu 38:

Tìm vi phân của các hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maakaaabaGaaG4maiaadIhacqGHRaWkcaaIYaaaleqaaaaa!3B6B! y = \sqrt {3x + 2} \)

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaiodaaeaadaGcaaqaaiaaiodacaWG4bGa % ey4kaSIaaGOmaaWcbeaaaaGccaWGKbGaamiEaaaa!3F11! dy = \frac{3}{{\sqrt {3x + 2} }}dx\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaWaaOaaaeaacaaIZaGa % amiEaiabgUcaRiaaikdaaSqabaaaaOGaamizaiaadIhaaaa!3FCB! dy = \frac{1}{{2\sqrt {3x + 2} }}dx\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaaiodacaWG4bGa % ey4kaSIaaGOmaaWcbeaaaaGccaWGKbGaamiEaaaa!3F0F! dy = \frac{1}{{\sqrt {3x + 2} }}dx\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpdaWcaaqaaiaaiodaaeaacaaIYaWaaOaaaeaacaaIZaGa % amiEaiabgUcaRiaaikdaaSqabaaaaOGaamizaiaadIhaaaa!3FCD! dy = \frac{3}{{2\sqrt {3x + 2} }}dx\)
Câu 39:

Tìm vi phân của các hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaIYaGaamiE % amaaCaaaleqabaGaaGOmaaaaaaa!3D6D! y = {x^3} + 2{x^2}\)

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHsislcaaI0aGaamiEaiaacMcacaWGKbGaamiEaaaa!4175! dy = (3{x^2} - 4x)dx\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHRaWkcaWG4bGaaiykaiaadsgacaWG4baaaa!40AC! dy = (3{x^2} + x)dx\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHRaWkcaaIYaGaamiEaiaacMcacaWGKbGaamiEaaaa!4168! dy = (3{x^2} + 2x)dx\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadM % hacqGH9aqpcaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHRaWkcaaI0aGaamiEaiaacMcacaWGKbGaamiEaaaa!416A! dy = (3{x^2} + 4x)dx\)
Câu 40:

Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaqa % daqaaiaadIhacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabe % aacaaIYaaaaaaa!4186! y = f\left( x \right) = {\left( {x - 1} \right)^2}\) . Biểu thức nào sau đây chỉ vi phân của hàm số f(x)?

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcaaIYaWaaeWaaeaacaWG4bGaeyOeI0IaaGymaaGaayjk % aiaawMcaaiaabsgacaWG4baaaa!3FAD! {\rm{d}}y = 2\left( {x - 1} \right){\rm{d}}x\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpdaqadaqaaiaadIhacqGHsislcaaIXaaacaGLOaGaayzk % aaWaaWbaaSqabeaacaaIYaaaaOGaaeizaiaadIhaaaa!3FE4! {\rm{d}}y = {\left( {x - 1} \right)^2}{\rm{d}}x\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcaaIYaWaaeWaaeaacaWG4bGaeyOeI0IaaGymaaGaayjk % aiaawMcaaaaa!3DC9! {\rm{d}}y = 2\left( {x - 1} \right)\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadM % hacqGH9aqpcaaIYaWaaeWaaeaacaWG4bGaeyOeI0IaaGymaaGaayjk % aiaawMcaaiaabsgacaWG4baaaa!3FAD! {\rm{d}}y = 2\left( {x - 1} \right){\rm{d}}x\)
Câu 41:

Cho chuyển động thẳng xác định bởi phương trình \(s=t^{3}-3 t^{2}\) (t tính bằng giây; s tính bằng mét). Khằng định nào sau đây đúng ?

A. Gia tốc của chuyền động khi t=4 s là \(a=18 m / s^{2}\)

B. Gia tốc của chuyền động khi t=4 s là \(a=9 m / s^{2}\)

C. Vận tốc của chuyền động khi t=3 s là v=12 m / s

D. Vận tốc của chuyền động khi t=3 s là v=24 m / s
Câu 42:

Một chuyền động thẳng xác định bởi phương trình \(s=t^{3}-3 t^{2}+5 t+2,\) trong dó t tính bằng giây và s tính bằng mét. Gia tốc của chuyển động khi t=3 là:

A. \(24 m / s^{2}\)

B. \(17 m / s^{2}\)

C. \(14 m / s^{2}\)

D. \(12 m / s^{2}\)
Câu 43:

Cho hàm số \(f(x)=\sqrt{\cos 2 x} .\) Khi đó

A. \(\mathrm{d}[f(x)]=\frac{\sin 2 x}{2 \sqrt{\cos 2 x}} \mathrm{d} x\)

B. \(\mathrm{d}[f(x)]=\frac{\sin 2 x}{\sqrt{\cos 2 x}} \mathrm{d} x\)

C. \(\mathrm{d}[f(x)]=\frac{-\sin 2 x}{2 \sqrt{\cos 2 x}} \mathrm{d} x\)

D. \(\mathrm{d}[f(x)]=\frac{-\sin 2 x}{\sqrt{\cos 2 x}} \mathrm{d} x\)
Câu 44:

Cho hàm số \(y=\frac{1-x^{2}}{1+x^{2}} .\) Vi phân của hàm số là:

A. \(\mathrm{d} y=\frac{-4 x}{\left(1+x^{2}\right)^{2}}\)

B. \( \mathrm{d} y=\frac{-4}{\left(1+x^{2}\right)^{2}} \mathrm{d} x\)

C. \( \mathrm{d} y=\frac{-4}{1+x^{2}} \mathrm{d} x\)

D. \(\mathrm{d} y=\frac{-\mathrm{d} x}{\left(1+x^{2}\right)^{2}}\)
Câu 45:

Vi phân của hàm số \(y=\frac{2 x+3}{2 x-1}\) là :

A. \(\mathrm{d} y=-\frac{8}{(2 x-1)^{2}} \mathrm{d} x\)

B. \(\mathrm{d} y=\frac{4}{(2 x-1)^{2}} \mathrm{d} x\)

C. \(\mathrm{d} y=-\frac{4}{(2 x-1)^{2}} \mathrm{d} x\)

D. \(\mathrm{d} y=-\frac{7}{(2 x-1)^{2}} \mathrm{d} x\)
Câu 46:

Cho hàm số \(y=f(x)=\sqrt{1+\cos ^{2} 2 x}\). Chọn kết quả đúng:

A. \(\mathrm{d} f(x)=\frac{-\sin 4 x}{2 \sqrt{1+\cos ^{2} 2 x}} \mathrm{d} x\)

B. \(\mathrm{d} f(x)=\frac{-\sin 4 x}{\sqrt{1+\cos ^{2} 2 x}} \mathrm{d} x\)

C. \(\mathrm{d} f(x)=\frac{\cos 2 x}{\sqrt{1+\cos ^{2} 2 x}} \mathrm{d} x\)

D. \(\mathrm{d} f(x)=\frac{-\sin 2 x}{\sqrt{1+\cos ^{2} 2 x}} \mathrm{d} x\)
Câu 47:

Cho hàm số \(f(x)=\left\{\begin{array}{l}x^{2}+x \text { khi } x \geq 0 \\ x \text { khi } x<0\end{array}\right.\). Khẳng định nào dưới đây là sai?

A. \(f^{\prime}\left(0^{+}\right)=1\)

B. \(f^{\prime}\left(0^{-}\right)=1\)

C. \(\mathrm{d} f(0)=\mathrm{d} x\)

D. Hàm số không có vi phân tại x=0
Câu 48:

Cho hàm số \(y=\cos ^{2} 2 x .\) Vi phân của hàm số là:

A. \(\mathrm{d} y=4 \cos 2 x \sin 2 x \mathrm{d} x\)

B. \(\mathrm{d} y=-2 \cos 2 x \sin 2 x \mathrm{d} x\)

C. \(\mathrm{d} y=2 \cos 2 x \sin 2 x \mathrm{d} x\)

D. \(\mathrm{d} y=-2 \sin 4 x \mathrm{d} x\)
Câu 49:

Cho hàm số \(f(x)=\left\{\begin{array}{lll}x^{2}-x & \text { khi } & x \geq0 \\ 2 x & \text { khi } & x<0\end{array}\right.\). Kết quả nào dưới đây đúng?

A. \(\mathrm{d} f(0)=-\mathrm{d} x\)

B. \(f^{\prime}\left(0^{+}\right)=\lim\limits _{x \rightarrow 0^{+}} \frac{x^{2}-x}{x}=\lim\limits _{x \rightarrow 0^{+}}(x-1)=-1\)

C. \(f^{\prime}\left(0^{+}\right)=\lim\limits _{x \rightarrow 0^{+}}\left(x^{2}-x\right)=0\)

D. \(f^{\prime}\left(0^{-}\right)=\lim\limits _{x \rightarrow 0^{-}} 2 x=0\)
Câu 50:

Cho hàm số \(y=\sin (\sin x) .\) Vi phân của hàm số là:

A. \(\mathrm{d} y=\cos (\sin x) \cdot \sin x \mathrm{d} x\)

B. \(\mathrm{d} y=\sin (\cos x) \mathrm{d} x\)

C. \(\mathrm{d} y=\cos (\sin x) \cdot \cos x \mathrm{d} x\)

D. \(\mathrm{d} y=\cos (\sin x) \mathrm{d} x\)

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Trắc nghiệm Vi phân Toán Lớp 11

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