\(x\left(x-y\right)+2\left(y-x\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)

\(=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)

\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1

Ta có lược đồ sau :

Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)

Bài 2:

1)  \(x^2-4x+4=\left(x-2\right)^2\)

2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)

3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)

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

5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

Bài 2:

7) \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

8) \(x^3+64=\left(x+4\right)\left(x^2+4x+16\right)\)

9) \(\left(a+b\right)^2-\left(2a-b\right)^2=\left(a+b+2a-b\right)\left(a+b-2a+b\right)=3a\left(-a+2b\right)\)

10) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a\cdot2b=4ab\)

11) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

\(=2a\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)

\(=2a\left(3a^2+b^2\right)\)

12) \(\left(6x-1\right)^2-\left(3x+2\right)^2=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)=\left(9x+1\right)\left(3x-3\right)\)

Ta có: \(y^2\left(x^2+y\right)-zx^2-zy\)

\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)

\(=\left(x^2+y\right)\left(y^2-z\right)\)

\(=\left(x+2\right)\left(5x-3-2x\right)\)

\(x\left(x+y\right)-6x-6y\)

\(=x\left(x+y\right)-6\left(x+y\right)\)

\(=\left(x-6\right)\left(x+y\right)\)

ai giải gimf nha

phần này học từ kì 1 rồi mà bn

\(-\left(x+2y\right)^2\)

= \(-\left(x^2+4xy+4y^2\right)\)

= \(-\left(x+2y\right)^2\)