Đề thi cuối kì 2 Toán 12 Chân trời sáng tạo – Đề 10 (có đáp án)

$a-b$.

$b-a$.

$a+b$.

$-a-b$.

$S=\underset{0}{\overset{2}{\int }}{3}^{x}dx$.

$S=\pi \underset{0}{\overset{2}{\int }}{3}^{2x}dx$.

$S=\pi \underset{0}{\overset{2}{\int }}{3}^{x}dx$.

$S=\underset{0}{\overset{2}{\int }}{3}^{2x}dx$.

$$M\left(-1;0;0\right)$$

$N\left(0;-2;0\right)$.

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

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

$\overrightarrow{u_2}=\left(3;4;-1\right)$.

$\overrightarrow{u_1}=\left(2;-5;2\right)$.

$\overrightarrow{u_3}=\left(2;5;-2\right)$.

$\overrightarrow{u_4}=\left(3;4;1\right)$.

$$0^{\circ }$$.

$90^{\circ }$.

$3^{\circ }$.

$45^{\circ }$.

$$\overrightarrow{n}=\left(0;1;1\right)$$.

$$\overrightarrow{n}=\left(0;-1;1\right)$$.

$$\overrightarrow{n}=\left(1;0;1\right)$$.

$$\overrightarrow{n}=\left(0;2;1\right)$$.

$\frac{x-1}{4}=\frac{y}{2}=\frac{z-5}{5}.$

$\frac{x+1}{4}=\frac{y}{2}=\frac{z+5}{5}.$

$\frac{x-4}{4}=\frac{y-2}{2}=\frac{z-5}{5}.$

$\frac{x-1}{5}=\frac{y}{2}=\frac{z-5}{4}.$

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

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

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

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

$M\left(1;2;3\right)$.

$Q\left(0;1;2\right)$.

$P\left(-1;1;-1\right)$.

$N\left(0;1;1\right)$.

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

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

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

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

$$A|B=\left\{\left(1,1\right);\left(1,3\right);\left(1,5\right);\left(3,1\right);\left(3,3\right);\left(3,5\right);\left(5,1\right);\left(5,3\right);\left(5,5\right)\right\}$$.

$$\begin{array}{lc}A|B=\{\left(1,1\right);\left(1,3\right);\left(1,5\right);\left(3,1\right);\left(3,3\right);\left(3,5\right);\left(5,1\right);\left(5,3\right);\left(5,5\right);& \\ & \left(1,2\right);\left(1,4\right);\left(1,6\right);\left(3,2\right);\left(3,4\right);\left(3,6\right);\left(5,2\right);\left(5,4\right);\left(5,6\right)\}\end{array}$$.

$$A|B=\left\{\left(1,2\right);\left(1,4\right);\left(1,6\right);\left(3,2\right);\left(3,4\right);\left(3,6\right);\left(5,2\right);\left(5,4\right);\left(5,6\right)\right\}$$.

$$A|B=\left\{\left(1,1\right);\left(1,3\right);\left(1,5\right);\left(3,3\right);\left(3,5\right);\left(5,1\right);\left(5,3\right);\left(5,5\right)\right\}$$.

$P\left(A\right)=P\left(B\right).P\left(A|B\right)+P\left(\bar{B}\right).P\left(A|\bar{B}\right).$

$P\left(A\right)=P\left(B\right).P\left(B|A\right)+P\left(\bar{B}\right).P\left(B|\bar{A}\right).$

$P\left(A\right)=P\left(A\right).P\left(A|B\right)+P\left(\bar{A}\right).P\left(A|\bar{B}\right).$

$P\left(A\right)=P\left(A\right).P\left(B|A\right)+P\left(\bar{A}\right).P\left(B|\bar{A}\right).$