Question

若非零有理数x，y，z满足x+y+z=0，求证√1/x²+1/y²+1/z²为有理数

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Answer 1

呵呵,没人解答我来吧

首先将 跟号下的式子 化简 ,分母是x²y²z²,很明显,分母是可以开根号的,开出根号是|xyz|

再看分子 化简后 分子是y²z²+x²y²+x²z²

由x+y+z=0 两边平方x²+y²+z²+2xy+2xz+2yz=0

xy+xz+yz=-(x²+y²+z²)/2

两边平方(xy+xz+yz)²=y²z²+x²y²+x²z²+2x²yz+2xy²z+2xyz²=y²z²+x²y²+x²z²+2xyz(x+y+z)

由于x+y+z=0 所以2xyz(x+y+z)=0

即(xy+xz+yz)²=y²z²+x²y²+x²z²

所以分子开根号即为|xy+xz+yz|

所得结果即(xy+xz+yz)/xyz的绝对值(根号下式子为非负数)

Answer 2

=√(1/x²+1/y²+1/(x+y)²)
=√(((x²+y²)(x+y)²+x²y²)/(x²y²(x+y)²))
=√(((x²+y²)²+2xy(x²+y²)+x²y²)/(x²y²(x+y)²))
=((x²+y²)²+2xy(x²+y²)+x²y²)/(x²y²(x+y)²)
又x,y为有理数，故该数为有理数