\(2xy-x^2-y^2+16\)

\(=-\left(x^2-2xy+y^2-16\right)\)

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

2x ( x - y ) - 10y ( x - y )

= ( x - y ) ( 2x - 10y )

= ( x - y ) ( 2 . x - 2 . 5y )

= ( x - y ) ( x - 5y ) 2

2x ( x - y ) - 10y ( x - y )

= ( x - y ) ( 2x - 10y )

= ( x - y ) ( 2 . x - 2 . 5y )

= ( x - y ) ( x - 5y ) 2

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

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

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

ta có :

\(x^2-5x+5y-y^2=\left(x^2-y^2\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-5\right)\)

x²-5x+5y-y²

=(x²-y²)-(5x-5y)

=(x-y)(x+y)-5(x-y)

=(x-y)(x+y-5)

x²-8x-15=x²-8x+16-31=(x-4)²-31=\(\left(x-4-\sqrt{31}\right).\left(x-4+\sqrt{31}\right)\)

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

x-y²+x²-y

=(x-y)+(x²-y²)

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

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

Lời giải:

$(x^3-2xy+3x^2)(x-2y)=x(x^2-2xy+3x)(x-2y)$

x² - 4x -5

= x² - 5x + x -5

= x(x-5) + (x-5)

=(x+1)(x-5)

x^2-4x-4=x²-2.2.x-2²

           =[x-2]²

\(x^2-4x-5=x^2-5x+x-5=\left(x^2-5x\right)+\left(x-5\right)=x\left(x-5\right)+\left(x-5\right)=\left(x+1\right)\left(x-5\right)\)

x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung