Đề giữa kì 1 Toán 11 năm 2025 – 2026 trường THPT Trà Ôn – Vĩnh Long

8; 16; 32; 65.

-1; -2; ; -4; -8.

1; -2; 11; -24.

4; 6; 8; 10.

$x = \dfrac{2\pi}{9} + k2\pi,x = - \dfrac{\pi}{6} + k2\pi$ $(k \in {\mathbb{Z}})$.

$x = \dfrac{\pi}{6} + k\pi,x = - \dfrac{\pi}{6} + k\dfrac{\pi}{3}$ $(k \in {\mathbb{Z}})$.

$x = \dfrac{\pi}{6} + k\dfrac{\pi}{6},x = - \dfrac{\pi}{6} + k\dfrac{\pi}{2}$ $(k \in {\mathbb{Z}})$.

$x = \dfrac{2\pi}{9} + k\pi,x = - \dfrac{\pi}{6} + k\dfrac{\pi}{3}$ $(k \in {\mathbb{Z}})$.

S = 59050.

S = -118100.

S = 118098.

S = -59048.

$\sin\left( {a - b} \right) = \sin a - \sin b$.

$\cos\left( {a + b} \right) = \cos a\cos b + \sin a\sin b$.

$\cos\left( {a + b} \right) = \cos a\cos b - \sin a\sin b$.

$\tan\left( {a + b} \right) = \tan a + \tan b$.

47,5.

44.

51.

23,75.

$\dfrac{2959\pi}{90}$.

$\dfrac{269\pi}{9}$.

$\dfrac{269\pi}{27}$.

$\dfrac{269\pi}{18}$.

$\dfrac{25}{8}$.

$\dfrac{37}{12}$.

$\dfrac{61}{8}$.

$\dfrac{49}{9}$.

$- 60{^\circ} + k360{^\circ}\left( {k \in {\mathbb{Z}}} \right)$.

$60{^\circ} + k180{^\circ}\left( {k \in {\mathbb{Z}}} \right)$.

$60{^\circ} + k2\pi\left( {k \in {\mathbb{Z}}} \right)$.

$60{^\circ} + k360{^\circ}\left( {k \in {\mathbb{Z}}} \right)$.

$x = \dfrac{\pi}{6} + k2\pi,\,\, k \in {\mathbb{Z}}$.

$x = \pm \dfrac{\pi}{6} + k\pi,\,\, k \in {\mathbb{Z}}$.

$x = \dfrac{5\pi}{6} + k\pi,\,\, k \in {\mathbb{Z}}$.

$x = \dfrac{\pi}{6} + k\pi,\,\, k \in {\mathbb{Z}}$.

$- \dfrac{\sqrt{2}}{3} + \dfrac{\sqrt{3}}{6}$.

$\dfrac{\sqrt{3}}{6} + \dfrac{\sqrt{2}}{3}$.

$\dfrac{1}{3} + \dfrac{2\sqrt{2}}{3}$.

$- \dfrac{1}{6} + \dfrac{\sqrt{6}}{3}$.

$S_{7} = 322$.

$S_{7} = 112$.

$S_{7} = 406$.

$S_{7} = \dfrac{721}{2}$.

14.

7.

$\dfrac{3}{10}$.

-13.