Đề thi Học kì 2 Toán 10 năm 2024-2025 trường Lê Thánh Tông (Tp.HCM)

$(2;4)$

$(5;6)$

$(15;10)$

$(50;6)$

40

391

780

1560

14

45

15

50

$(x+1)^2+(y+3)^2=3$

$(x-1)^2+(y-3)^2=3$

$(x+1)^2+(y+3)^2=\sqrt{3}$

$(x-1)^2+(y-3)^2=\sqrt{3}$

6

72

720

144

2023

2024

2025

2026

$\left\{ \begin{array}{l} x=5+5t \\ y=-7t \end{array}\right.$

$\left\{ \begin{array}{l} x=5+5t \\ y=7t \end{array}\right.$

$\left\{ \begin{array}{l} x=5+5t \\ y=1+7t \end{array}\right.$

$\left\{ \begin{array}{l} x=5-5t \\ y=2-7t \end{array}\right.$

$S=\varnothing$

$S = \{6\}$

$S = \{3\}$

$S = \{6; 3\}$

$\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$

$\dfrac{y^2}{16} + \dfrac{x^2}{9} = 1$

$\dfrac{y^2}{16} - \dfrac{x^2}{9} = 1$

$\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$

$F_1(-3;0); F_2(3;0)$

$F_1(0; -\sqrt{8}); F_2(0; \sqrt{8})$

$F_1(-5;0); F_2(5;0)$

$F_1(-\sqrt{8};0); F_2(\sqrt{8};0)$

$m > \dfrac{3}{2}$

$m > \dfrac{3}{4}$

$\dfrac{3}{4} < m < \dfrac{3}{2}$

$1 < m < 3$