Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2x^2+2y^2-x^2z+z-y^2z-2\)
\(=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)\)
\(=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)\)
\(=\left(2-z\right)\left(x^2+y^2-1\right)\)
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
\(a,x^2-x-y^2+y\\=(x^2-y^2)-(x-y)\\=(x-y)(x+y)-(x-y)\\=(x-y)(x+y-1)\\---\\b,x^2+2x+2z-z^2\\=(x^2-z^2)+(2x+2z)\\=(x-z)(x+z)+2(x+z)\\=(x+z)(x-z+2)\\\text{#}Toru\)
Lời giải:
a. $x^2-x-y^2+y=(x^2-y^2)-(x-y)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1)$
b. $x^2+2x+2z-z^2=(x^2+2x+1)-(z^2-2z+1)=(x+1)^2-(z-1)^2$
$=(x+1-z+1)(x+1+z-1)=(x-z+2)(x+z)$
=xy ( x + y ) + z ( x^2 + 2xy + y^2 ) = xy ( x + y ) + z ( x + y ) ^ 2 = ( x + y ) ( xy + xz + yz )
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left[\left(z-y\right)+\left(y-x\right)\right]\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-y\right)-x^2z^2\left(y-x\right)\)
\(=\left(y-x\right)\left(x^2y^2-x^2z^2\right)+\left(z-y\right)\left(y^2z^2-x^2z^2\right)\)
\(=x^2\left(y-x\right)\left(y-z\right)\left(y+z\right)+z^2\left(z-y\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(z-y\right)\left(-x^2y-x^2z+z^2y+z^2x\right)\)
\(=\left(y-x\right)\left(z-y\right)\left[xz\left(z-x\right)+y\left(z-x\right)\left(z+x\right)\right]\)
\(=\left(y-x\right)\left(z-y\right)\left(z-x\right)\left(xy+yz+xz\right)\)
y(x−2z)2+8xyz+x(y−2z)2−2z(x+y)2
=y(x2−4xz+4z2)+8xyz+x(y2−4yz+4z2)−2z(x2+y2+2xy)
=(yx2+xy2)+(4yz2+4xz2)−(2zx2+2zy2+4xyz)
=xy(x+y)+4z2(x+y)−2z(x2+y2+2xy)
=xy(x+y)+4z2(x+y)−2z(x+y)2
=(x+y)(xy+4z2−2xz−2yz)
=(x+y)[y(x−2z)−2z(x−2z)]
=(x+y)(y−2z)(x−2z)