Đề thi Học kì 1 Toán 10 năm 2024-2025 trường THPT Bình Chiểu (Tp.HCM)

$c^2 = a^2 + b^2 - 2ab \cos A$

$c^2 = a^2 + b^2 + 2ab \cos A$

$c^2 = a^2 + b^2 - 2ab \cos C$

$c^2 = a^2 + b^2 + 2ab \cos C$

$a = 2R \sin A$

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

$R = 2a \sin A$

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

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

$S = pr$

$S = \sqrt{p(p - a)(p - b)(p - c)}$

$S = \dfrac{a + b + c}{4R}$

$\sqrt{19}$

$2\sqrt{7}$

$3 \sqrt{19}$

$2 \sqrt{19}$

$\overrightarrow{OM} - \overrightarrow{ON} = \overrightarrow{MN}$

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

$\overrightarrow{AB} + \overrightarrow{CA} = \overrightarrow{CB}$

$\overrightarrow{AM} - \overrightarrow{MN} = \overrightarrow{AN}$

$\overrightarrow{AG} + \overrightarrow{BG} + \overrightarrow{CG} = \overrightarrow{0}$

$\overrightarrow{GA} + \overrightarrow{GB} = \overrightarrow{GC}$

$\overrightarrow{GA} + \overrightarrow{GB} = \overrightarrow{CG}$

$\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = \overrightarrow{0}$

$\vec{a} \cdot \vec{b} = |\vec{a}| . |\vec{b}|$

$\vec{a} \cdot \vec{b} = |\vec{a}| . |\vec{b}| . \cos \left( \vec{a}; \vec{b} \right)$

$\vec{a} \cdot \vec{b} = \left| \vec{a} \vec{b} \right| . \cos \left( \vec{a}; \vec{b} \right)$

$\vec{a} \cdot \vec{b} = |\vec{a}| . |\vec{b}| . \sin \left( \vec{a}; \vec{b} \right)$

$\overrightarrow{AB}.\overrightarrow{AD} = 0$

$\overrightarrow{AB}.\overrightarrow{AD} = a$

$\overrightarrow{AB}.\overrightarrow{AD} = \dfrac{a^2}{2}$

$\overrightarrow{AB}.\overrightarrow{AD} = a^2$

$2a$

$a$

$a\sqrt{3}$

$2\sqrt{3}a$

3

38

41

19