\(x^2-x-xy-2y^2+2y\)

\(=x^2-x-2xy+xy-2y^2+2y\)

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

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

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

Mk lm câu a nha! :D 

a) 4x + 3x - 10 

 = ( 4x - 5 ) ( x+2 ) 

^^ Học tốt!  

Dúng không thế I have a crazy idea

Làm luôn câu b đi I have a crazy idea

x³-3x²-4=

Bạn ơi ở câu hỏi tương tự có pp bậc ba trở lên đấy

=8x^2+12x=4x(2x+3)

\(x^2+7x^2+12x=8x^2+12x=4x\left(2x+3\right)\)

a, = (x^3-x^2)-(4x^2-4x)+(4x-4)

    = (x-1).(x^2-4x+4) = (x-1).(x-2)^2

b, = (x^3+x^2)-(10x^2+10x)+(16x+16)

 = (x+1).(x^2-10x+16)

 = (x+1).[ (x^2-2x)-(8x-16) ] = (x+1).(x-2).(x-8)

k mk nha

a)= (x^3-x^2)-(4x^2-4x)+(4x-4)

    = (x-1).(x^2-4x+4)

    = (x-1).(x-2)^2

b)= (x^3+x^2)-(10x^2+10x)+(16x+16)

 = (x+1).(x^2-10x+16)

 = (x+1).[ (x^2-2x)-(8x-16) ]

 = (x+1).(x-2).(x-8)

P/s tham khảo nha

\(x^3+3x^2-4\)

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

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

Mình nhìn nhầm đề

\(x^3+3x^2-4\)

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

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

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

\(=\left(x+2\right)\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)

\(=\left(x+2\right)\left(x+2\right)\left(x-1\right)\)

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

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

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