已知a(a+1)-(a²+b)=-2,求2分之a²+b²减ab的值
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已知a(a+1)-(a²+b)=-2,求2分之a²+b²减ab的值
已知a(a+1)-(a²+b)=-2,求2分之a²+b²减ab的值
已知a(a+1)-(a²+b)=-2,求2分之a²+b²减ab的值
a²+a-a²-b=-2
a-b=-2
所以原式=(a²+b²-2ab)/2
=(a-b)²/2
=4/2
=2
a(a+1)-(a²+b)=-2
a²+a-a²-b=-2
a-b=2
2分之a²+b²减ab
=1/2(a-b)²
=1/2*4
=2
2