已知a²+b²+2a+4b+5=0.求代数式[（a+b/2)+(a-b/2)²]·（2a&

a^2+b^2+2a+4b+5=0
a^2+2a+1+b^2+4b+4=0
(a+1)^2+(b+2)^2=0
所以a+1=0,b+2=0a=-1,b=-2.
[（a+b/2）+(a-b/2)^2]*(2a^2-1/2b^2)=[(-1-1)+(-1+1)^2]*(2-1/2*4)
=-2*0
=0