当a=1,b=-2时,求[(a+½b)²+(a-½b)²](2a²-½b²)的值
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当a=1,b=-2时,求[(a+½b)²+(a-½b)²](2a²-½b²)的值
当a=1,b=-2时,求[(a+½b)²+(a-½b)²](2a²-½b²)的值
当a=1,b=-2时,求[(a+½b)²+(a-½b)²](2a²-½b²)的值
=[(1-2)^2+(1+2)^2]*[2*1^2-1/2*(-2)^2]=(1+9)*(2-1/2*4)=10*(2-2)=0