Đề thi giữa học kì 2 Toán 10 - Đề số 4

4.

Không xác định.

16.

3.

$\text{D} = {\mathbb{R}}\backslash\left\{ 3 \right\}.$

$\text{D} = \left\lbrack {1; + \infty} \right).$

$\text{D} = {\mathbb{R}}.$

$\text{D} = {\mathbb{R}}\backslash\left\{ 1 \right\}.$

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

$y = \dfrac{2x}{x + 2}$.

$y = - 2x^{2} + 3x + 1$.

$y = 3x + 5$.

$I\left( {- \dfrac{1}{3};\dfrac{2}{3}} \right)$.

$I\left( {\dfrac{1}{3};\dfrac{2}{3}} \right)$.

$I\left( {\dfrac{1}{3}; - \dfrac{2}{3}} \right)$.

$I\left( {\dfrac{1}{3};\dfrac{3}{2}} \right)$.

$x \in \left( {- \infty\,;\, 0} \right) \cup \left( {2\,;\, + \infty} \right)$.

$x \in \left( {0\,;\, 2} \right)$.

$x \in \left( {2\,;\, + \infty} \right)$.

$x \in {\mathbb{R}}$.

$S = \left( {- \,\infty; - 6} \right\rbrack.$

$S = \left\lbrack {1; + \,\infty} \right).$

$S = \left\lbrack {- 6;1} \right\rbrack.$

$S = \left( {- \,\infty; - 6} \right\rbrack \cup \left\lbrack {1; + \,\infty} \right).$

S = {2}.

S = {3}.

S = {1}.

S = {-1}.

$\overset{\rightarrow}{n} = ( - 7;4)$.

$\overset{\rightarrow}{n} = (4; - 7)$.

$\overset{\rightarrow}{n} = (7;4)$.

$\overset{\rightarrow}{n} = (4;7)$.

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

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

$\left\{ \begin{array}{l} {x = 5 + 4t} \\ {y = 6 - 3t} \end{array} \right.$.

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

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$5.$

$2.$

$3.$

$4.$