Question

已知a^2+b^2+2a+4b+5=0 求代数式 [（a+b/2）+(a-b/2)^2]*(2a^2-1/2b^2)的植

Answer 1

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=0

a=-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