一道与二次根式有关的问题

题目

一道与二次根式有关的问题
已知2006x²=2007y²=2008z² ,
√2006x+2007y+2008z=√2006+√2007+√2008
求1/x+1/y+1/z

答案

令2006x²=2007y²=2008z²=k则
2006x²=k x=√k/2006 y=√k/2007 z=√k/2008
√2006x+2007y+2008z=3√k=√2006+√2007+√2008
k= (√2006+√2007+√2008 )^2/9
2006x²=k= (√2006+√2007+√2008 )^2/9
x=(√2006+√2007+√2008)/(3√2006)
y=(√2006+√2007+√2008)/(3√2007)
z=(√2006+√2007+√2008)/(3√2008)
1/x+1/y+1/z =3(√2006+√2007+√2008)/(√2006+√2007+√2008)
=3