\(\left(2a-b-2\right)\left(a+3b+1\right)\)

Bạn ơi làm thế nào để ra kết quả đấy thế?

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) x² + 4x + 3 - y² -2y

= x² +4x + 4 - y² -2y-1

= (x+2)² - (y+1)²

= (x+2-y-1).(x+2+y+1)

= (x-y+1).(x+y+3)

b) 2a² -5ab + 2b²

= 2a² -4ab + 2b² - ab

= 2.(a² - 2ab+b²) - ab

= 2.(a-b)² -ab

...

c) (x+y)² - 2x - 2y + 1

= (x+y)² - 1 - 2x -2y +2

= (x+y-1).(x+y+1) - 2.(x+y-1)

= (x+y-1)²

c: =5a(a-b)+7(a-b)

=(a-b)(5a+7)

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

\(=xy\left[\left(x-y\right)\left(x+y\right)+3\left(x-y\right)\right]\)

\(=xy\left(x-y\right)\left(x+y+3\right)\)

\(Ht\)

nếu sai cho mik xl vì mik chx thành thục cái này

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

1: \(6x^2y-9xy^2+3xy\)

\(=3xy\left(2x-3y+1\right)\)

2: \(\left(4-x\right)^2-16\)

\(=\left(4-x-4\right)\left(4-x+4\right)\)

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

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

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

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

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)