已知实数a,b满足(a+b)²=1,(a-b)²=5,求a²+b+ab的值?
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已知实数a,b满足(a+b)²=1,(a-b)²=5,求a²+b+ab的值?
已知实数a,b满足(a+b)²=1,(a-b)²=5,求a²+b+ab的值?
已知实数a,b满足(a+b)²=1,(a-b)²=5,求a²+b+ab的值?
题目应是:已知实数a,b满足(a+b)²=1,(a-b)²=5,a²+b²+ab的值
(a+b)^2=1
a^2+2ab+b^2=1
(a-b)^2=5
a^2-2ab+b^2=5
两式相加2a^2+2b^2=6
a^2+b^2=3
两式相减4ab=-4
ab=-1
a^2+b^2+ab=3-1=2
是a²+b平方+ab吧