Đề giữa kỳ 2 Toán 10 năm 2024 – 2025 trường THPT số 1 TX Sa Pa – Lào Cai

${\mathbb{R}}\backslash\left\{ 1 \right\}$.

${\mathbb{R}}\backslash\left\{ {0;3} \right\}$.

$\mathbb{R}$.

$\left( {0;3} \right)$.

$d_{({M;\Delta})} = \dfrac{ax_{0} + by_{0} + c}{\sqrt{a^{2} + b^{2} + c^{2}}}$.

$d_{({M;\Delta})} = \dfrac{\left| {ax_{0} + by_{0} + c} \right|}{\sqrt{a^{2} + b^{2} + c^{2}}}$.

$d_{({M;\Delta})} = \dfrac{ax_{0} + by_{0} + c}{\sqrt{a^{2} + b^{2}}}$.

$d_{({M;\Delta})} = \dfrac{\left| {ax_{0} + by_{0} + c} \right|}{\sqrt{a^{2} + b^{2}}}$.

(3; 0).

(0; 3).

(1; 3).

(3; 1).

$\mathbb{R}$.

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

$\left( {- 2; + \infty} \right)$.

${\mathbb{R}}\backslash\left\{ 2 \right\}$.

$f(x) = 3x^{2} + 2x - 5$ là tam thức bậc hai.

$f(x) = 2x - 4$ là tam thức bậc hai.

$f(x) = 3x^{3} + 2x - 1$ là tam thức bậc hai.

$f(x) = x^{4} - x^{2} + 1$ là tam thức bậc hai.

$3x^{2} - 4x + 1 = {(x - 2)}^{2}.$

$3x^{2} - 4x + 1 = (x - 2).$

$3x^{2} - 4x + 1 = - 9{(x - 2)}^{2}.$

$3x^{2} - 4x + 1 = 3{(x - 2)}^{2}.$

I(-2; -3).

I(2; 3).

I(3; -2).

I(3; 2).

$\left( {x - 3} \right)^{2} + \left( {y - 1} \right)^{2} = 25$.

$\left( {x + 3} \right)^{2} + \left( {y - 1} \right)^{2} = 5$.

$\left( {x - 3} \right)^{2} + \left( {y - 1} \right)^{2} = 5$.

$\left( {x + 3} \right)^{2} + \left( {y + 1} \right)^{2} = 25$.

$\left( {- \dfrac{b}{2a};\mkern6mu \mkern6mu \dfrac{\Delta}{4a}} \right).$

$\left( {\dfrac{b}{2a};\mkern6mu \mkern6mu \dfrac{\Delta}{4a}} \right).$

$\left( {- \dfrac{b}{a};\mkern6mu - \mkern6mu \dfrac{\Delta}{4a}} \right).$

$\left( {- \dfrac{b}{2a};\mkern6mu - \mkern6mu \dfrac{\Delta}{4a}} \right).$

$\overset{\rightarrow}{u_{1}} = \left( {2\,;\, 3} \right).$

$\overset{\rightarrow}{u_{2}} = \left( {3\,;\, - 1} \right).$

$\overset{\rightarrow}{u_{3}} = \left( {1\,;\, 3} \right).$

$\overset{\rightarrow}{u_{4}} = \left( {- 3\,;\, 1} \right).$

$\left( {- \infty\,;\, - 3} \right\rbrack \cup \left\lbrack {4\,;\, + \infty} \right)$.

$\varnothing$.

$\left( {- \infty\,;\, - 4} \right\rbrack \cup \left\lbrack {3\,;\, + \infty} \right)$.

$\left\lbrack {- 3\,;\, 4} \right\rbrack$.

${\mathbb{R}}\backslash\left\{ {\pm 3} \right\}$.

${\mathbb{R}}\backslash\left\{ {- 3} \right\}$.

${\mathbb{R}}\backslash\left\{ 3 \right\}$.

$\left( {3; + \infty} \right)$