\(a,x^2-x-y^2+y\\=(x^2-y^2)-(x-y)\\=(x-y)(x+y)-(x-y)\\=(x-y)(x+y-1)\\---\\b,x^2+2x+2z-z^2\\=(x^2-z^2)+(2x+2z)\\=(x-z)(x+z)+2(x+z)\\=(x+z)(x-z+2)\\\text{#}Toru\)
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. $x^2+2x+2z-z^2=(x^2+2x+1)-(z^2-2z+1)=(x+1)^2-(z-1)^2$
$=(x+1-z+1)(x+1+z-1)=(x-z+2)(x+z)$
\(a,x^2-x-y^2+y\\ =\left(x^2-y^2\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y-1\right)\\ ---\\ b,x^2+2x+2z-z^2\\ =\left(x^2+2x+1\right)-\left(z^2-2z+1\right)\\ =\left(x+1\right)^2-\left(z-1\right)^2\\ =\left[\left(x+1\right)+\left(z-1\right)\right].\left[\left(x+1\right)-\left(z-1\right)\right]\\ =\left(x+z\right)\left(x-z+2\right)\)
