先分解因式后计算求值（1）(a+b/2)²-(a-b/2)²,a=-1/8,b=2
问题描述：
先分解因式后计算求值（1）(a+b/2)²-(a-b/2)²,a=-1/8,b=2
（2）x²-16/8-2x,其中x= -2 已知x+y=1,求1/2x²+xy+1/2y²的值
答
(a+b/2)²-(a-b/2)²,
=(a+b/2+a-b/2)(a+b/2-a+b/2)
=2axb
=2ab
=2x(-1/8)x2
=-1/2
x²-16/8-2x
=(x+4)(x-4)/2(4-x)
=-(x+4)/2
=-(-2+4)/2
=-1
1/2x²+xy+1/2y²
=1/2(x+y)²
=1/2x1²
=1/2