Bộ 24 Đề kiểm tra Giữa kì 2 Toán 11 có đáp án (Mới nhất) (Đề 14)

a = -5;

a = 1;

a = 4;

a = 5

$\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{BC};$

$\overrightarrow{AB}+\overrightarrow{CA}=\overrightarrow{BC};$

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

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

$\overrightarrow{CB},\overrightarrow{AC},\overrightarrow{BE};$

$\overrightarrow{EF},\overrightarrow{FH},\overrightarrow{EG};$

$\overrightarrow{AB},\overrightarrow{BC},\overrightarrow{AE};$

$\overrightarrow{DB},\overrightarrow{AC},\overrightarrow{DF}.$

a²;

$\frac{a^2\sqrt{2}}{2};$

$a^2\sqrt{3};$

$a^2\sqrt{2}.$

$\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AC}=\overrightarrow{AC\text{'}};$

$\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA\text{'}}=\overrightarrow{AC};$

$\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA\text{'}}=\overrightarrow{AC\text{'}};$

$\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA\text{'}}=\overrightarrow{BC}.$

0° < φ < 180°;

0° £ φ £ 90°;

0° < φ < 90°;

0° £ φ £ 180

$\overrightarrow{u}.\overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos \left(\overrightarrow{v},\overrightarrow{u}\right);$

$\overrightarrow{u}.\overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\sin \left(\overrightarrow{u},\overrightarrow{v}\right);$

$\overrightarrow{u}.\overrightarrow{v}=\overrightarrow{u}.\overrightarrow{v}\cos \left(\overrightarrow{u},\overrightarrow{v}\right);$

$\overrightarrow{u}.\overrightarrow{v}=\overrightarrow{u}.\overrightarrow{v}\tan \left(\overrightarrow{u},\overrightarrow{v}\right).$

0

-7;

9;

-9

y - y₀ = f (x₀)(x - x₀);

$y=f^{\text{'}}\left(x_0\right)\left(x+x_0\right)+y_0;$

$y+y_0=f^{\text{'}}\left(x_0\right)\left(x-x_0\right);$

$y=f^{\text{'}}\left(x_0\right)\left(x-x_0\right)+y_0.$

1

$\frac{3}{2};$

$\frac{1}{9};$

$\frac{3}{4};$

$\underset{x\to 1^{+}}{\lim }y=3;$

$\underset{x\to 1^{+}}{\lim }y=0;$

$\underset{x\to 1^{+}}{\lim }y=-\infty ;$

$\underset{x\to 1^{+}}{\lim }y=+\infty .$

góc giữa hai vectơ chỉ phương của chúng là 90°;

tích vô hướng giữa hai vectơ chỉ phương của chúng là bằng 0;

góc giữa hai vectơ chỉ phương của chúng là 0°;

góc giữa hai đường thẳng đó là 90°.

$\lim {\left(\frac{1}{3}\right)}^n=3;$

$\lim \frac{2}{n+2}=2;$

$\lim \frac{1}{n}=1;$

$\lim \frac{1}{n}=0.$

-3

$-\frac{1}{3};$

2

$\frac{1}{2}.$

y = tan x;

y = -x³ + 3x² - x + 1;

$y=\frac{x+1}{x-2};$

y = cot x.

45°;

60°;

90°;

120°.

3;

0;

4;

6

60°;

90°;

120°;

45°.

$\underset{x\to 1}{\lim }5=1;$

$\underset{x\to 2}{\lim }\left(2x\right)=2;$

$\underset{x\to 2}{\lim }{x}^4=-\infty ;$

$\underset{x\to +\infty }{\lim }{x}^3=+\infty .$

$y=\frac{x+1}{x-2};$

$y=\frac{x-1}{x};$

$y=\frac{x-2}{x+1};$

$y=\frac{x+1}{x+2}.$