Họ tất cả các nguyên hàm của hàm f ( x ) = e 5 x f\left(x\right)=e^{5x} f ( x ) = e 5 x là:

1 5 e 5 x + C \frac{1}{5}e^{5x}+C 5 1 e 5 x + C

e 5 x . ln ⁡ 5 + C e^{5x}.\ln5+C e 5 x . ln 5 + C

5 e 5 x + C 5e^{5x}+C 5 e 5 x + C

Thể tích vật thể tròn xoay khi hình phẳng giới hạn bởi y = x 3 -1, trục Ox, trục Oy quay quanh Ox là:

9 π 14 \frac{9\pi}{14} 1 4 9 π

14 π 9 \frac{14\pi}{9} 9 1 4 π

d : x − 3 = y + 1 2 = z − 2 4 d:\frac{x}{-3}=\frac{y+1}{2}=\frac{z-2}{4} d : − 3 x = 2 y + 1 = 4 z − 2 vec tơ nào dưới đây là vec tơ chỉ phương của đường thẳng

a → = ( 6 ; − 4 ; − 8 ) \overrightarrow{a}=\left(6;-4;-8\right) a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 6 ; − 4 ; − 8 )

a → = ( 0 ; − 1 ; 2 ) \overrightarrow{a}=\left(0;-1;2\right) a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 0 ; − 1 ; 2 )

a → = ( 0 ; 1 ; − 2 ) \overrightarrow{a}=\left(0;1;-2\right) a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 0 ; 1 ; − 2 )

a → = ( 3 ; − 2 ; 4 ) \overrightarrow{a}=\left(3;-2;4\right) a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 3 ; − 2 ; 4 )

Vec tơ nào sau đây là vec tơ pháp tuyến của mặt phẳng (P): x - 2y +3 = 0

n → = ( − 1 ; 2 ; 0 ) \overrightarrow{n}=\left(-1;2;0\right) n <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( − 1 ; 2 ; 0 )

n → = ( − 1 ; 2 ; 3 ) \overrightarrow{n}=\left(-1;2;3\right) n <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( − 1 ; 2 ; 3 )

n → = ( 1 ; − 2 ; 3 ) \overrightarrow{n}=\left(1;-2;3\right) n <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 1 ; − 2 ; 3 )

n → = ( 1 ; 2 ; 0 ) \overrightarrow{n}=\left(1;2;0\right) n <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 1 ; 2 ; 0 )

đường thẳng d qua A(1;2;-3) và vuông góc với mặt phẳng (P): x - 2y +z -4=0 có phương trình là:

x = 2 − t ; y = 2 t ; z = − 2 − t x=2-t;\ y=2t;\ z=-2-t x = 2 − t ; y = 2 t ; z = − 2 − t

x + 1 1 = y + 2 − 2 = z − 3 1 \frac{x+1}{1}=\frac{y+2}{-2}=\frac{z-3}{1} 1 x + 1 = − 2 y + 2 = 1 z − 3

x − 1 1 = y − 2 2 = z + 3 1 \frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1} 1 x − 1 = 2 y − 2 = 1 z + 3

x = 1 − t ; y = 2 − 2 t ; z = − 3 + t x=1-t;\ y=2-2t;\ z=-3+t x = 1 − t ; y = 2 − 2 t ; z = − 3 + t

khoảng cách từ A(-1;3;4) đến mặt phẳng (P): x + 3y +z - 1 = 0 bằng

11 \sqrt{11} 1 1 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

13 11 \frac{13}{\sqrt{11}} 1 1 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> 1 3

12 11 \frac{12}{\sqrt{11}} 1 1 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> 1 2

11 11 \frac{\sqrt{11}}{11} 1 1 1 1 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Tích có hướng của hai vec tơ $$\overrightarrow{a\ }$$ và $$\overrightarrow{b\ }$$ là một vectơ:

vuông góc với cả hai vectơ đó.

cùng phương với cả hai vectơ đó.

đồng phẳng với cả hai vectơ đó.

bằng tích vô hướng của hai vectơ đó.

untitled

Quiz

•

Mathematics

•

12th Grade

•

Practice Problem

•

Medium

Trang Quỳnh

Used 1+ times

FREE Resource

36 questions

Show all answers

1.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Tập M có 6 phần tử. Số tập con gồm 2 phần tử của M là?

15

30

2

$2^6$

2.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Cho cấp số cộng $\left(u_n\right)$ có $u_1=-2;d=3$ . Tìm $u_{10}?$

$u_{10}=28$

$u_{10}=-29$

$u_{10}=25$

$u_{10}=-2.3^9$

3.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Cho hàm số f(x) có bảng biến thiên . Hàm số f(x) đồng biến trên khoảng nào sau?

(2;5)

(-2;2)

(-2;0)

(0;2)

4.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Cho hàm số f(x) có bản biến thiên sau. Điểm cực đại của hàm số đã cho là

$x=-3$

$x=1$

$x=2$

$x=-2$

5.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Cho hàm số f(x) có đồ thị f'(x) như hình vẽ. Hàm số $y=f\left(x\right)$ có bao nhiêu cực trị?

3

1

2

4

6.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Đường cong ở hình bên là đồ thị của hàm nào?

$y=\frac{2x-1}{x+1}$

$y=\frac{2x+1}{x-1}$

$y=\frac{2x+1}{x+1}$

$y=\frac{1-2x}{x-1}$

7.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Cho $\log_ax=-1;\log_ay=4$ . Giá trị của $\log_a\left(x^2y^3\right)$

3

-14

65

10

8.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Nghiệm của phương trình $\log_{ }\left(x-1\right)=2\$ là:

$x=3$

$x=5$

$x=101$

$x=-3$

9.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Họ tất cả các nguyên hàm của hàm $f\left(x\right)=e^{5x}$ là:

$e^{5x}.\ln5+C$

$e^{5x}+C$

$5e^{5x}+C$

$\frac{1}{5}e^{5x}+C$

10.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Nếu $\int_0^2f\left(x\right)dx=3.\int_0^2\left[4f\left(x\right)-3\right]$ bằng:

8

2

6

4

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untitled

36 questions

Tập M có 6 phần tử. Số tập con gồm 2 phần tử của M là?

Cho cấp số cộng (un)\left(u_n\right)(un​) có u1=−2;d=3u_1=-2;d=3u1​=−2;d=3 . Tìm u10?u_{10}?u10​?

Cho hàm số f(x) có bảng biến thiên . Hàm số f(x) đồng biến trên khoảng nào sau?

Cho hàm số f(x) có bản biến thiên sau. Điểm cực đại của hàm số đã cho là

Cho hàm số f(x) có đồ thị f'(x) như hình vẽ. Hàm số y=f(x)y=f\left(x\right)y=f(x) có bao nhiêu cực trị?

Đường cong ở hình bên là đồ thị của hàm nào?

Cho log⁡ax=−1;log⁡ay=4\log_ax=-1;\log_ay=4loga​x=−1;loga​y=4 . Giá trị của log⁡a(x2y3)\log_a\left(x^2y^3\right)loga​(x2y3)

Nghiệm của phương trình log⁡(x−1)=2 \log_{ }\left(x-1\right)=2\ log​(x−1)=2 là:

Họ tất cả các nguyên hàm của hàm f(x)=e5xf\left(x\right)=e^{5x}f(x)=e5x là:

Nếu ∫02f(x)dx=3.∫02[4f(x)−3]\int_0^2f\left(x\right)dx=3.\int_0^2\left[4f\left(x\right)-3\right]∫02​f(x)dx=3.∫02​[4f(x)−3] bằng:

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Cho cấp số cộng $\left(u_n\right)$ có $u_1=-2;d=3$ . Tìm $u_{10}?$

Cho hàm số f(x) có đồ thị f'(x) như hình vẽ. Hàm số $y=f\left(x\right)$ có bao nhiêu cực trị?

Cho $\log_ax=-1;\log_ay=4$ . Giá trị của $\log_a\left(x^2y^3\right)$

Nghiệm của phương trình $\log_{ }\left(x-1\right)=2\$ là:

Họ tất cả các nguyên hàm của hàm $f\left(x\right)=e^{5x}$ là:

Nếu $\int_0^2f\left(x\right)dx=3.\int_0^2\left[4f\left(x\right)-3\right]$ bằng: