\(1,\\ a,x=\dfrac{\sqrt{3}-\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}=0\\ \Leftrightarrow A=\left(0-0-1\right)^2+2021=1+2021=2022\\ b,\left(x+\sqrt{x^2+2021}\right)\left(y+\sqrt{y^2+2021}\right)=2021\\ \Leftrightarrow\left(x-\sqrt{x^2+2021}\right)\left(x+\sqrt{x^2+2021}\right)\left(y+\sqrt{y^2+2021}\right)=2021\left(x-\sqrt{x^2+2021}\right)\\ \Leftrightarrow-2021\left(y+\sqrt{y^2+2021}\right)=2021\left(x-\sqrt{x^2+2021}\right)\)

Cmttt \(\Leftrightarrow-2021\left(x+\sqrt{x^2+2021}\right)=2021\left(y-\sqrt{y^2+2021}\right)\)

Cộng vế theo vế

\(\Leftrightarrow-2021y-2021\sqrt{y^2+2021}-2021x-2021\sqrt{x^2+2021}=2021x-2021\sqrt{x^2+2021}+2021y-2021\sqrt{y^2+2021}\\ \Leftrightarrow x+y=0\\ \Leftrightarrow x=-y\\ \Leftrightarrow x^{2021}+y^{2021}=x^{2021}-x^{2021}=0\)

c - d - a -b

C

D

A

B

C

D

A

B

Bài 3: 

g: \(\left|x\right|< 2\)

nên -2<x<2

i: Ta có: \(\left|x-3\right|=x-3\)

\(\Leftrightarrow x-3\ge0\)

hay \(x\ge3\)

1)

a) 4y²-4xy+x²= x²-4xy+4y²= (x-2y)²

b) 9x²-12xy+4y²= (3x)²-2.3x.2y+(2y)²= (3x-2y)²

c) 16x²-25=(4x)²-5²= (4x-5)(4x+5)

d) 1-9y²= 1²-(3y)²=(1-3y)(1+3y)

g) x³-27y³= (x-3y)(x²+3xy+9y²)

h) 64 + 8x³=(4+2x)(16+8x+4x²)

i) x⁴+4y⁴-4x²y²= (x²)²-2x²2y²+(y²)²=(x²-y²)²

1 c

1C

2 ko có đáp án đúng

3B

4A

5C

6D

7D

8B

Bài 3:

\(1,x=9\Leftrightarrow A=\dfrac{3-2}{9+3}=\dfrac{1}{12}\\ 2,P=AB=\dfrac{\sqrt{x}-2}{x+3}\cdot\dfrac{x-3\sqrt{x}+2-2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x+3\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{x+3}\\ 3,\left(10x+30\right)P\ge x+25\\ \Leftrightarrow\dfrac{3\sqrt{x}\left(x+3\right)}{x+3}-x-25\ge0\\ \Leftrightarrow3\sqrt{x}-x-25\ge0\\ \Leftrightarrow-\left(x-3\sqrt{x}+\dfrac{9}{4}\right)-\dfrac{91}{4}\ge0\\ \Leftrightarrow-\left(\sqrt{x}-\dfrac{3}{2}\right)^2-\dfrac{91}{4}\ge0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\)

khuyến cáo ko nên gạt xuống.

Đồ ngu đồ ăn hại cút mịa mài đê :D