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已知正整数x,y,z,w满足2007x^2=2008y^2=2009z^2=2010w^2,且1/x + 1/y + 1/z+ 1/w =1.(问题见下)
求根号下2007x+2008y+2009z+2010w的值.
人气:312 ℃ 时间:2019-11-10 19:08:00
解答
2007x^2=2008y^2=2009z^2=2010w^2 = k所以2007x = k/x,2008y= k/y,2009z = k/z,2010w =k/w,所以√(2007x) + √(2008y)+√(2009z)+√(2010w) = √k (1/√x+1/√y+1/√z+1/√w)=√k(1/√(k/2007) + 1/√(k/2008) + 1/...
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