已知a2+b2+2a+4b+5=0.求代数式[(a+b/2)+(a-b/2)2]*(2a2-1/2b2)的值?

a²+b²+2a+4b+5=0
(a²+2a+1)+(b²+4b+4)=0
(a+1)²+(b+2)²=0
a+1=0,b+2=0
a=-1,b=-2
[(a+b/2)+(a+b/2)²]×(2a²-1/2b²)
=[(a+b/2)(1+a+b/2)]×2(a²-1/4b²)
=2(a+b/2)(1+a+b/2)(a+b/2)(a-b/2)
=2(a+b/2)²(a-b/2)(1+a+b/2)
=2×[-1+(-2)/2]²(-1+2/2)(1-1-2/2)
=2×(-2)²×0×(-1)
=0