Đề thi giữa học kì 1 Toán 10 - Đề số 4

“$\forall x \in \mathbb{R},{x^2} - x + 1 \ge 0$”

“$\forall x \in \mathbb{R},{x^2} - x + 1 \lt 0$”

“$\forall x \in \mathbb{R},{x^2} - x + 1 \gt 0$”

“$\exists x \in \mathbb{R},{x^2} - x + 1 \ge 0$”

$x + 4{y^3} \gt 0$

$x + y = 3$

$x + \frac{1}{{{y^2}}} \ge 0$

$x + \frac{y}{2} \lt 0$

$x - 2y - 2 \gt 0$

$5x - 2y - 2 \gt 0$

$5x - 2y - 1 \gt 0$

$4x - 2y - 2 \gt 0$

O(0;0)

M(1;2)

N(2;1)

P(8;4)

${a^2} = {b^2} + {c^2} + 2bc\cos \widehat {BAC}$

${a^2} = {b^2} + {c^2} - 2bc\sin \widehat {BAC}$

${a^2} = {b^2} + {c^2} + 2bc\sin \widehat {BAC}$

${a^2} = {b^2} + {c^2} - 2bc\cos \widehat {BAC}$

[-2;8]

($- \infty$;8]

{-2}

$\emptyset$

$\left\{ {\begin{array}{cccccccccccccccccccc}{x - y \ge 0}\\{2x - y \ge 1}\end{array}} \right.$

$\left\{ {\begin{array}{cccccccccccccccccccc}{x - y \gt 0}\\{2x - y \gt 1}\end{array}} \right.$

$\left\{ {\begin{array}{cccccccccccccccccccc}{x - y \lt 0}\\{2x - y \gt 1}\end{array}} \right.$

$\left\{ {\begin{array}{cccccccccccccccccccc}{x - y \lt 0}\\{2x - y \lt 1}\end{array}} \right.$

$\sin ({90^o} - \alpha ) = - \cos \alpha$

$\sin ({90^o} - \alpha ) = - \sin \alpha$

$\sin ({90^o} - \alpha ) = \cos \alpha$

$\sin ({90^o} - \alpha ) = \sin \alpha$

$\forall x \in ( - 4;1] \Leftrightarrow - 4 \lt x \lt 1$

$\forall x \in ( - 4;1] \Leftrightarrow - 4 \le x \lt 1$

$\forall x \in ( - 4;1] \Leftrightarrow - 4 \lt x \le 1$

$\forall x \in ( - 4;1] \Leftrightarrow - 4 \le x \le 1$

${S_{\Delta ABC}} = \frac{{7\sqrt 3 }}{2}$

${S_{\Delta ABC}} = 14$

${S_{\Delta ABC}} = 7$

${S_{\Delta ABC}} = 7\sqrt 3$

$\{ 1;3\}$

$\{ 2;5\}$

$\{ 0;3\}$

$\{ 0;1;2\}$

$\tan \alpha \gt 0$

$\sin \alpha \lt 0$

$\cos \alpha \gt 0$

$\cot \alpha \lt 0$