Dạng 1

Đại số

Câu 1. B

Câu 2. B

Câu 3. D

Câu 4. B

Câu 5. B

Câu 6. C

Hình học

Câu 1. A

Câu 2. D

Câu 3. A

Câu 4. B

Câu 5. D

Câu 6. D

Hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

Cau 23: B

Câu 24: A

23 là b

còn 24 là a

Câu 8: A

hong hiểu

8 là a

Câu 13: D

Câu 14: C

15A 

mí câu nì k bt lm thì thế số vô tính cx đựt :>

Lời giải:

Kẻ $OH\perp AB$ thì $OH=1$ (cm)

Áp dụng định lý Pitago cho tam giác $OHA$ vuông:

$AH=\sqrt{OA^2-OH^2}=\sqrt{3^2-1^2}=2\sqrt{2}$ (cm)

$OA=OB$ nên tam giác $OAB$ cân tại $O$. Do đó đường cao $OH$ đồng thời là đường trung tuyến 

$\Rightarrow AB=2AH=4\sqrt{2}$ (cm)

Hình vẽ:

Lời giải:

HPT \(\Leftrightarrow \left\{\begin{matrix} x=-3|y|\\ 3x+y=-3\end{matrix}\right.\)

\(\Rightarrow -9|y|+y=-3\)

Nếu $y\geq 0$ thì pt trở thành:

$-9y+y=-3$

$\Leftrightarrow y=\frac{3}{8}$

$x=-3|y|=-3.\frac{3}{8}=\frac{-9}{8}$

Nếu $y< 0$ thì pt trở thành:

$9y+y=-3\Leftrightarrow y=\frac{-3}{10}$

$x=-3|y|=-3.\frac{3}{10}=\frac{-9}{10}$

Vậy hpt có 2 nghiệm $(x,y)=(\frac{3}{8}, \frac{-9}{8}); (\frac{-3}{10}, \frac{-9}{10})$