Bài `1`

\(a,A=a\left(a+b\right)-b\left(a+b\right)\\ =\left(a+b\right)\left(a-b\right)\)

Với `a=9;=10`

Ta có :

 \(\left(a+b\right)\left(a-b\right)\\=\left(9+10\right)\left(9-10\right)\\ =19.\left(-1\right)\\ =-19\)

\(b,B=\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(3x+2\right)\left(3x-2\right)\\ =\left(3x+2\right)^2-2\left(3x+2\right)\left(3x-2\right)+\left(3x-2\right)^2\\ =\left[\left(3x+2\right)-\left(3x-2\right)\right]^2\)

Với `x=-4`

Ta có :

\(\left[\left(3x+2\right)-\left(3x-2\right)\right]^2\\ =\left(3.4+2-3.4+2\right)^2\\ =\left(12+2-12+2\right)^2\\ =4^2\\ =16\)

\(2,\\ x^3-6x^2+9x\\ =x\left(x^2-6x+9\right)\\ =x\left(x-3\right)^2\\ x^2-2x-4y^2-4y\\ \)

`->` có đúng đề ko cậu

2:

b; x^2-4y^2-2x-4y

=(x-2y)*(x+2y)-2(x+2y)

=(x+2y)(x-2y-2)

a: x^3-6x^2+9x

=x(x^2-6x+9)

=x(x-3)^2

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VIII
1. too
2. so
3. doesn't
4. can't
5. too

VII
1A
2A
3D
4D
5A

1B
2A
3B
4C
5C
6D
7C
8C
9B
10B
11C
12A
13D
14A
15C
16D
17B
18C
19C
20D
21D
22D