初三上学期

每日一题：2020-09-30

每日一题: 2020-09-30

题目: x,y,zx,y,z 为正实数, 且满足xyz=1,x+1z=5,y+1x=29xyz=1,x+\frac{1}{z}=5,y+\frac{1}{x}=29, 求z+1yz+\frac{1}{y} 的值.

参考思路

分析: 考虑x+1z,y+1x,z+1yx+\frac{1}{z},y+\frac{1}{x},z+\frac{1}{y} 的乘积, 化不对称为对称.
5⋅29⋅(z+1y)=(x+1z)(y+1x)(x+1y)=xyz+x+y+z+1x+1y+1z+1xyz5\cdot 29\cdot (z+\frac{1}{y})=(x+\frac{1}{z})(y+\frac{1}{x})(x+\frac{1}{y})=xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}
=1+(x+1z)+(y+1x)+(z+1y)+1=36+(z+1y)=1+(x+\frac{1}{z})+(y+\frac{1}{x})+(z+\frac{1}{y})+1=36+(z+\frac{1}{y})
所以可得z+1y=14z+\frac{1}{y}=\frac{1}{4}.