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设0<b<a,a2+b2-6ab=0,则a+b/a-b
人气:180 ℃ 时间:2020-01-11 23:26:32
解答
∵a>b>0∴a²+b²-6ab=0等式两边同除以b²,得:∴(a/b)²+1-6(a/b)=0记t=a/b【t>1】,则上式可化为:t²-6t+1=0△=36-4=32∴t=(6+4√2)/2=3+2√2,即:a/b=3+2√2∴(a+b)/(a-b)=(a/b+1)/...
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