已知a2+b2+2a+4b+5=0.求代数式[(a+b/2)+(a-b/2)2]*(2a2-1/2b2)的值?
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已知a2+b2+2a+4b+5=0.求代数式[(a+b/2)+(a-b/2)2]*(2a2-1/2b2)的值?
![已知a2+b2+2a+4b+5=0.求代数式[(a+b/2)+(a-b/2)2]*(2a2-1/2b2)的值?](/uploads/image/z/3953820-12-0.jpg?t=%E5%B7%B2%E7%9F%A5a2%2Bb2%2B2a%2B4b%2B5%3D0.%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F%5B%28a%2Bb%2F2%29%2B%28a-b%2F2%292%5D%2A%282a2-1%2F2b2%29%E7%9A%84%E5%80%BC%3F)
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
(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
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