分解因式:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z,其中x=‐1,y=1/2,z=‐3/4

Question

分解因式:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z,其中x=‐1,y=1/2,z=‐3/4

Answer 1

:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z
=(2x²-x²+y²)(z²-x²-y²+z²)+z
=(x²+y²)[2z²-(x²+y²)]+z
=2z²(x²+y²)-(x²+y²)²+z
当x=‐1,y=1/2,z=‐3/4时,上式得：
2*(-3/4)²[(-1)²+(1/2)²]-[(-1)²+(1/2)²]²+(-3/4)
=2*9/16(1+1/4)-(1+1/4)²-3/4
=9/8*5/4-20/16-3/4
=1/32(45-40-24)
=-19/32