先因式分解再求值(2分之x+y)^2-(2分之x-y)^2,其中x=-8分之1,y=2
问题描述：
先因式分解再求值
(2分之x+y)^2-(2分之x-y)^2,其中x=-8分之1,y=2
答
∵x=-1/8,y=2
∴[(x+y)/2]²-[(x-y)/2]²
=(x+y)²/4-(x-y)²/4
=1/4·[(x+y)²-(x-y)²]
=1/4×[(x+y)+(x-y)][(x+y)-(x-y)]
=1/4×(x+y+x-y)(x+y-x+y)
=1/4×2x×2y
=xy
=(-1/8)×2
=-1/4