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

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

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

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

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

I(-2; -3).

I(2; 3).

I(3; -2).

I(3; 2).

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

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

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

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

\(\left( {- \dfrac{b}{2a};\mspace{6mu}\mspace{6mu}\dfrac{\Delta}{4a}} \right).\)

\(\left( {\dfrac{b}{2a};\mspace{6mu}\mspace{6mu}\dfrac{\Delta}{4a}} \right).\)

\(\left( {- \dfrac{b}{a};\mspace{6mu} - \mspace{6mu}\dfrac{\Delta}{4a}} \right).\)

\(\left( {- \dfrac{b}{2a};\mspace{6mu} - \mspace{6mu}\dfrac{\Delta}{4a}} \right).\)

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

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

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

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

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

$\varnothing$.

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

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

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

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

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

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