已知(a+b)²=m,(a-b)²=n,求:(1)(a²+b²)/ab;(2)ab的值用m,n来表示
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已知(a+b)²=m,(a-b)²=n,求:(1)(a²+b²)/ab;(2)ab的值用m,n来表示
已知(a+b)²=m,(a-b)²=n,求:(1)(a²+b²)/ab;(2)ab的值
用m,n来表示
已知(a+b)²=m,(a-b)²=n,求:(1)(a²+b²)/ab;(2)ab的值用m,n来表示
(1)2(m+n)/(m-n)
(2) (m-n)/4
(a+b)²=m,(a-b)²=n,
即:a^2+2ab+b^2=m
a^2-2ab+b^2=n
两式相加得 2(a^2+b^2)=m+n
两式相减得4ab=m-n
所以
(1)(a²+b²)/ab=2*(m+n)/(m-n)
(2)ab=(m-n)/4
解由(a+b)²=m
得a^2+b^2+2ab=m...............①
又由,(a-b)²=n
得a^2+b^2-2ab=n...............②
两式相加得2(a^2+b^2)=m+n
即(a^2+b^2)=(m+n)/2
由①-②得
4ab=m-n
即ab=(m-n)/4
故(a²+b²)/ab
=[(m+n)/2]/[(m-n)/4]
=2(m+n)/(m-n)
ab=(m-n)/4