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数学题:难题
设2006x3=2007y3=2008z3,且xyz>0,3√2006x2+2007y2+2008z2=3√2006+3√2007+3√2008,求1/x+1/y+1/z
人气:270 ℃ 时间:2020-02-23 14:04:52
解答
设2006x^3=2007y^3=2008z^3=k(2006x2+2007y2+2008z2)^(1/3)=2006^(1/3)+2007^(1/3)+2008^(1/3)==>(k/x+k/y+k/z)^(1/3)=(k/x^3)^(1/3)+(k/y^3)^(1/3)+(k/z^3)^(1/3)==>k^(1/3)(1/x+1/y+1/z)^(1/3)=k^(1/3)(1/x+1/y+1/...
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