把下列各式分解因式: (1)x2-y2-z2+2yz; (2)(x+y)2+4(x+y+1)

问题描述:

把下列各式分解因式:
(1)x2-y2-z2+2yz;
(2)(x+y)2+4(x+y+1)

(1)x2-y2-z2+2yz
=x2-(y2+z2-2yz)
=x2-(y-z)2
=(x+y-z)(x-y+z);

(2)(x+y)2+4(x+y+1)
=(x+y)2+4(x+y)+4
=(x+y+2)2