\(a,5x^3y-10x^2y^2\\=5x^2y(x-2y)\\b,x^4-y^4\\=(x^2)^2-(y^2)^2\\=(x^2-y^2)(x^2+y^2)\\=(x-y)(x+y)(x^2+y^2)\)

\(c,(x+5)^2-16\\=(x+5)^2-4^2\\=(x+5-4)(x+5+4)\\=(x+1)(x+9)\\d,7x(y-3)-14(3-y)\\=7x(y-3)+14(y-3)\\=(7x+14)(y-3)\\=7(x+2)(y-3)\\Toru\)

A

D bạn nhé

a) 3(x-y) - (x-y)^2

 =(x-y)(3-x+y)

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

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

\(a,=3\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(3-x+y\right)\\ b,=\left(x+y\right)^2-4x^2y^2=\left(x-2xy+y\right)\left(x+2xy+y\right)\\ c,=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)

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

y²(\(x\)  + y) - ( \(x\) - 7)² (đk \(x\) +y ≥ 0)

= (y\(\sqrt{\left(x+y\right)}\) )² - (\(x\) - 7)²

= (y\(\sqrt{x+y}\) - (\(x-7\)))( y\(\sqrt{x+y}\) + (\(x\) - 7))

= (y\(\sqrt{x+y}\) - \(x\) + 7)(y\(\sqrt{x+y}\) + \(x\) - 7)

b) \(\left(a^2+b^2\right)^2-4a^2b^2\)

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

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

c) \(a^4-b^4=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)

a) \(y^4-y^3+y^2-y\)

\(=y^3\left(y-1\right)+y\left(y-1\right)\)

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

d) \(64m^3+8y^3=\left(4m+2y\right)\left(16m^2-8my+4y^2\right)\)

x . x   –   y   +   x 2 –   y 2

= x x − y + x − y x + y = 2 x + y x − y

Lời giải:

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

$=[(2x^2-xy)+(2xy-y^2)]-(2x-y)=[x(2x-y)+y(2x-y)]-(2x-y)$

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

\(a,=x\left(x-3\right)\\ b,=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\\ c,=\left(3x+7\right)\left(x-2\right)+2\left(x-2\right)=\left(x-2\right)\left(3x+9\right)=3\left(x+3\right)\left(x-2\right)\)