a) \(10x^2+10xy-x-y=10x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(10x-1\right)\)

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

\(2ab^2-a^2b-b^3\)

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

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

-(2ab² - a²b - b³)

= b(-2ab + a² + b²)

= b(a² - 2ab + b²)

= b(a - b)²

Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

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

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

\(=\left(2a-b\right).3b\)

\(=3b.\left(2a-b\right)\)

a) \(3a-3b+a^2-2ab+b^2\)

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

\(=\left(a-b\right)\left(a-b+3\right)\)

a)

3.(a-b) +2.(a-b ) =5 .(a-b )

câu b làm tương tự nha nhóm a^2 -2ab +b^2 vào 1nhoms và làm như câu a

mình làm nhầm 

a) \(\left(a-b\right).\left(a-b+3\right)\)

`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`

\(a,=x^2-9-x^2+6x-9=6x-18\\ b,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x+y-5\right)\left(x-y\right)\)

16ab + 4b² - 9 + 16a²

= (16a² + 16ab + 4b²) - 9

= (4a+2b)² - 3²

= (4a+2b-3)(4a+2b+3)