设a>b>0,求证:(a²-b²)/(a²+b²)>(a-b)/(a+b)
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设a>b>0,求证:(a²-b²)/(a²+b²)>(a-b)/(a+b)
设a>b>0,求证:(a²-b²)/(a²+b²)>(a-b)/(a+b)
设a>b>0,求证:(a²-b²)/(a²+b²)>(a-b)/(a+b)
先假设他俩相等,因为(A2-B2)=(a+b)*(a-b),等式两边均乘以[(a+b)/(a-b) ],之后化简得(a+b)2=a2+b2,然后就因为a>b>0,故2ab>0,得出题目中等式,得证.