Đề thi Giữa kì 1 Toán 10 năm 2024-2025 trường THPT Huỳnh Thúc Kháng (Nghệ An)

$\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{BC}$

$\mathbf{\overline{AB}} = \mathbf{\overline{DC}}$

$\overrightarrow{AC} = \overrightarrow{BA} - \overrightarrow{BC}$

$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$

$M = \left[-1;3 \right)$

$M = \left(-1;3 \right]$

$M = \left[-1;3 \right]$

$M = \left(-1;3 \right)$

$\sin\left(180^{\circ} - \alpha\right) = \sin \alpha$

$\cos\left(180^{\circ} - \alpha\right) = \cos \alpha$

$\cot\left(180^{\circ} - \alpha\right) = - \cot \alpha \left(0^{\circ} < \alpha < 180^{\circ} \right)$

$\tan\left(180^{\circ} - \alpha\right) = - \tan \alpha \left(\alpha \neq 90^{\circ}\right)$

$\exists x \in \mathbb{R}, x^{2} - x + 5 < 0$

$\exists x \in \mathbb{R}, x^{2} - x + 5 \ge 0$

$\forall x \in \mathbb{R}, x^{2} - x + 5 < 0$

$\forall x \in \mathbb{R}, x^{2} - x + 5 > 0$

$\cos \alpha = - \dfrac{2 \sqrt{2}}{3}$

$\cos \alpha = \dfrac{2}{3}$

$\cos \alpha = - \dfrac{2}{3}$

$\cos \alpha = \dfrac{2 \sqrt{2}}{3}$

$\dfrac{a}{\cos A} = 2R$

$S = \dfrac{abc}{4r}$

$S = \sqrt{p \left(p + a\right) \left(p + b\right) \left(p + c\right)}$

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

$\left\{\begin{array}{l} 3x+2y > 2 \\ x-y < 0 \end{array}\right.$

$\left\{\begin{array}{l} 7x-y < 1 \\ x + 6y^2 > 4y \end{array}\right.$

$\left\{\begin{array}{l} x^2 + y \geq 0 \\ x - y < 0 \end{array}\right.$

$\left\{\begin{array}{l} 2x^3 - y - 1 > 0 \\ x - 2 > 0 \\ x + 5y = 4 \end{array}\right.$

$x + 3y < 0$

$x - 3y > 0$

$-3x + 2y - 4 > 0$

$2x + y - 4 > 0$

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$2 + 2 = 5$