Bộ 15 đề thi giữa kì 2 Toán 12 có đáp án năm 2022-2023 (Đề 9)

$\int f\left(x\right)\text{d}x=f^{\text{'}}\left(x\right)+C$.

$\int f\left(x\right)\text{d}x=f^{\text{'}}\left(x\right)$.

$\int f^{\text{'}}\left(x\right)\text{d}x=f\left(x\right)+C$.

$\int f^{\text{'}}\left(x\right)\text{d}x=f\left(x\right)$.

$3x^2+3+C$.

$\frac{1}{4}{x}^4+\frac{3}{2}{x}^2+1+C$.

$\frac{1}{4}{x}^4+\frac{3}{2}{x}^2+x+C$.

$x^4+3x^2+x+C$.

$\frac{1}{2}\ln \left|2x+3\right|+2018$.

$\frac{1}{2}\ln \left|2x+3\right|-2018$.

$\ln \left|2x+3\right|+2018$.

$2\ln \left|2x+3\right|+2018$.

${\text{e}}^x.{\text{e}}^{x+1}+C$.

$\frac{1}{2}{\text{e}}^{2x+1}+C$.

$2{\text{e}}^{2x+1}+C$.

${\text{e}}^{x^2+x}+C$.

$\int f^{\text{'}}\left(x\right)\ln x\text{d}x=-\frac{1}{m}\left(\frac{2\ln x}{x^2}+\frac{1}{x^2}\right)+C$.

$\int f^{\text{'}}\left(x\right)\ln x\text{d}x=\frac{1}{m}\left(\frac{2\ln x}{x^2}+\frac{1}{x^2}\right)+C$.

$\int f^{\text{'}}\left(x\right)\ln x\text{d}x=-\frac{1}{m}\left(\frac{\ln x}{x^2}+\frac{1}{2x^2}\right)+C$.

$\int f^{\text{'}}\left(x\right)\ln x\text{d}x=-\frac{1}{m}\left(\frac{2\ln x}{x^2}-\frac{1}{x^2}\right)+C$.

-2.

-4.

2.

4.

ln 2.

$\ln 2-1$.

$1-\ln 2$.

$-\ln 2$.

$-2$.

2.

$-1$.

1.

$\left[\begin{array}{l}b=1\\ b=4\end{array}\right.$.

$\left[\begin{array}{l}b=0\\ b=2\end{array}\right.$.

$\left[\begin{array}{l}b=1\\ b=2\end{array}\right.$.

$\left[\begin{array}{l}b=0\\ b=4\end{array}\right.$.

$a+2b=0$.

$2a-b=0$.

$a-b=0$.

$a+b=0$.

$S=3$.

$S=-3$.

$S=5$.

$S=7$.

$V=\pi \underset{a}{\overset{b}{\int }}\left|f\left(x\right)\right|d\text{x}.$

$V=\underset{a}{\overset{b}{\int }}\left|f\left(x\right)\right|d\text{x}.$

$V=\pi \underset{a}{\overset{b}{\int }}{f}^2\left(x\right)d\text{x}.$

$V=\underset{a}{\overset{b}{\int }}{f}^2\left(x\right)d\text{x}.$

$\frac{27}{4}$.

$\frac{5}{6}$.

D. .

$\frac{24}{7}$.

$S=\frac{25}{6}.$

$S=\frac{20}{3}.$

$S=\frac{10}{3}.$

$S=9.$

$k={\text{e}}^2$.

$k=2e$.

$k=4$.

$k=8$.

$a\sqrt{5}$.

$5a^2$.

$a\sqrt{3}$.

$a\sqrt{2}$.

Số phức $z=i^2$ là số thuần ảo.

Số 3 không phải là số phức.

Số phức $z=3i+4$ có phần thực là 3 và phần ảo là 4.

Số phức liên hợp của $z=3i+4$ là $z=4-3i$.

$\left(3;4\right)$.

$\left(3;-4\right)$.

$\left(-3;-4\right)$.

$\left(-4;3\right)$.

$\overrightarrow{AB}=\left(x_1+x_2;y_1+y_2;z_1+z_2\right)$.

$\overrightarrow{AB}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$.

$\overrightarrow{AB}=\left(x_2-x_1;y_2-y_1;z_2-z_1\right)$.

$\overrightarrow{AB}=\left(x_1-x_2;y_1-y_2;z_1-z_2\right)$.

$D\left(1;1;2\right).$

$D\left(4;1;0\right).$

$D\left(-1;-1;-2\right).$

$D\left(-3;-1;0\right).$

$m=3$.

$m=2$.

$m=1$.

$m=0$.

3

4

2

1

$\overrightarrow{n}=\left(1;2;3\right).$

$\overrightarrow{n}=\left(1;-2;3\right).$

$\overrightarrow{n}=\left(1;3;-2\right).$

$\overrightarrow{n}=\left(1;-2;-3\right).$

$d\left(A,\left(P\right)\right)=\frac{12}{\sqrt{14}}$.

$d\left(A,\left(P\right)\right)=\frac{8}{\sqrt{14}}$.

$d\left(A,\left(P\right)\right)=\frac{1}{\sqrt{14}}$.

$d\left(A,\left(P\right)\right)=\frac{8}{\sqrt{6}}$.

$x-2y+3z=1.$

$\frac{x}{1}+\frac{y}{-2}+\frac{z}{3}=6.$

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

$6x-3y+2z=6.$

$\left(Q\right):2x+2y+3z-7=0.$

$\left(Q\right):2x-2y+3z-7=0.$

$\left(Q\right):2x+2y+3z-9=0.$

$\left(Q\right):x+2y+3z-7=0.$

$\left(d\right)\left\{\begin{array}{l}x=1-t\\ y=2+2t\\ z=-1+3t\end{array}\right.$.

$\left(d\right)\left\{\begin{array}{l}x=1+t\\ y=2-2t\\ z=-1+3t\end{array}\right.$.

$\left(d\right)\left\{\begin{array}{l}x=1+t\\ y=2+2t\\ z=1+3t\end{array}\right.$.

$\left(d\right):\left\{\begin{array}{l}x=1+t\\ y=2+2t\\ z=-1+3t\end{array}\right.$.

$\left\{\begin{array}{l}x=-4-2t\\ y=2-4t\\ z=-6-t\end{array}\right.$.

$\left\{\begin{array}{l}x=2-2t\\ y=1-4t\\ z=-3-t\end{array}\right.$.

$\left\{\begin{array}{l}x=2+2t\\ y=1+4t\\ z=-3+t\end{array}\right.$.

$\left\{\begin{array}{l}x=-4+2t\\ y=-2+4t\\ z=6+t\end{array}\right.$.

$\left\{\begin{array}{l}x=1+3t\\ y=-2+4t\\ z=3-7t\end{array}\right.$.

$\left\{\begin{array}{l}x=1+4t\\ y=-2+3t\\ z=3-7t\end{array}\right.$.

$\left\{\begin{array}{l}x=1+4t\\ y=2+3t\\ z=3-7t\end{array}\right.$.

$\left\{\begin{array}{l}x=1-4t\\ y=-2+3t\\ z=3-7t\end{array}\right.$.

$\frac{x-5}{6}=\frac{y-1}{5}=\frac{z-3}{3}$.

$\frac{x+5}{6}=\frac{y+1}{5}=\frac{z+3}{3}$.

$\frac{x-6}{5}=\frac{y-5}{1}=\frac{z-3}{3}$.

$\frac{x+6}{5}=\frac{y+5}{1}=\frac{z+3}{3}$.

$\frac{32}{3}\text{m}$.

$\frac{71}{3}\text{m}$.

$\frac{38}{3}\text{m}$.

$\frac{71}{6}\text{m}$.

0.

1.

2.

3.

$P=4$.

$P=\ln 2$.

$P=\ln 4041$.

$P=1$.

20.

8.

$\sqrt{46}$.

$2\sqrt{2}$.

$\frac{8\sqrt{3}}{3}$.

$\frac{8\sqrt{77}}{77}$.

$\frac{16\sqrt{77}}{77}$.

$\frac{16\sqrt{3}}{3}$.

${\left(x-4\right)}^2+y^2+z^2=14.$

${\left(x+4\right)}^2+y^2+z^2=14.$

$x^2+{(y-4)}^2+z^2=14.$

$x^2+y^2+{(z-4)}^2=14.$

${\left(x-1\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2=16.$

${\left(x-1\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2=9.$

${\left(x+1\right)}^2+{\left(y-2\right)}^2+{\left(z+3\right)}^2=16.$

${\left(x-1\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2=4.$

4

3

6

1

$\left\{\begin{array}{c}x=4t^{\text{'}}\\ y=-2-2t^{\text{'}}\\ z=-3\end{array}\right.$.

$\left\{\begin{array}{c}x=4t^{\text{'}}\\ y=2-2t^{\text{'}}\\ z=-3\end{array}\right.$.

$\left\{\begin{array}{c}x=4t^{\text{'}}\\ y=2+2t^{\text{'}}\\ z=-3\end{array}\right.$.

$\left\{\begin{array}{c}x=4t^{\text{'}}\\ y=2+2t^{\text{'}}\\ z=3\end{array}\right.$.