Đề thi Học kì 1 Toán 10 năm 2023-2024 trường THPT Thanh Đa (Tp.HCM)

$(-1;4]$

$\{0;1;2\}$

$\{1;2;3\}$

$\{0;1;2;3;4\}$

$(-1;2)$

$[-1;2]$

$(-1;2]$

$[-1;2)$

2

1

0

5

$(1;2)$

$(1;-2)$

$(0;-4)$

$(1;5)$

$f(-2) = 10$

$f\left(\dfrac{1}{5}\right) = -1.$

$f(-1) = 5.$

$f(2) = 10.$

$A(2;0)$

$B(1;-1)$

$C\left(3; \dfrac{1}{3}\right)$

$D(-1;-3)$

$\mathscr{D} = \mathbb{R} \setminus \{1\}$

$\mathscr{D} = [1; +\infty)$

$\mathscr{D} = \mathbb{R}$

$\mathscr{D} = (1; +\infty)$

$y = - \dfrac{3}{2}$

$x = - \dfrac{3}{2}$

$x = -3$

$y = -3$

$y = 2x^{2} - 4x - 3$

$y = 2x^{2} - 2x - 1$

$y = 2x^{2} + 4x + 5.$

$y = 2x^{2} + x + 2.$

360.

375.

370.

365.

$b\sin B = 2R$

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

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

$\sin C = \dfrac{c\sin A}{a}$

$AB^2 = AC^2 + BC^2 - 2AC.BC \cos C$

$AB^2 = AC^2 - BC^2 + 2AC.BC \cos C$

$AB^2 = AC^2 + BC^2 - 2AC.AB \cos C$

$AB^2 = AC^2 + BC^2 - 2AC.BC + \cos C$

5

$5\sqrt{3}$

10

$10\sqrt{3}$

$\overrightarrow{AB} = \overrightarrow{CD}$

$\overrightarrow{OA} = \overrightarrow{OC}$

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

$\overrightarrow{AD} = \overrightarrow{BC}$

$\overrightarrow{TA} = -\overrightarrow{TB}$

$\overrightarrow{TA} = \overrightarrow{TB}$

$\overrightarrow{AI} = \overrightarrow{BI}$

$IA = IB$

40

$-40$

13

$-13$