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=0a+1=0,b+2=0a=-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