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\(\left(x^3+3x^2y+3xy^2+y^3-z^3\right):\left(x+y-z\right)\\ =\left[\left(x+y\right)^3-z^3\right]:\left(x+y-z\right)\\ =\left(x+y-z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2\right]:\left(x+y-z\right)\\ =x^2+2xy+y^2+xz+yz+z^2\)
Vậy chọn A
xz-yz-x2+2xy-y2
=(xz-yz)-(x2-2xy+y2)
=z(x-y)-(x-y)2
=(x-y)(z-x+y)
=\(x^2+2xy+y^2-xz-zy\)
=\(\left(x+y\right)^2-z\left(x+y\right)\)
=\(\left(x+y\right)\left(x+y-z\right)\)
\(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
\(x^3-3x^2+3x-9=x^2\left(x-3\right)+3\left(x-3\right)=\left(x-3\right)\left(x^2+3\right)\)
\(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
\(x^2-2xy+y^2-yz+xz\)
\(=\left(x^2-2xy+y^2\right)+\left(xz-yz\right)\)
\(=\left(x-y\right)^2+z\left(x-y\right)\)
\(=\left(x-y+z\right)\left(x-y\right)\)
\(xz-yz-x^2+2xy-y^2\)
\(=z\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(z-x+y\right)\)
m)xz−yz−x2+2xy−y2
\(=z.\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
= \(z.\left(x-y\right)-\left(x-y\right)^2\)
= (x-y).(z-x+y)