题目

已知lgx=a,lgy=b,lgz=c,且a+b+c=0,求x^(1/b+1/c) y^(1/a+1/c) z^(1/b+1/a)的值

提问时间：2020-10-31

答案

a+b+c=0
a+b=-c b+c=-a a+c=-b
x^(1/b+1/c) y^(1/a+1/c) z^(1/b+1/a)=m (两边同时取对数)
lgm=lg[x^(1/b+1/c) y^(1/a+1/c) z^(1/b+1/a)]
=lg[x^(1/b+1/c) ]+lg[y^(1/a+1/c) ]+lg[z^(1/b+1/a)]
=(1/b+1/c)lgx+(1/a+1/c)lgy+(1/b+1/a)lgz 因为lgx=a,lgy=b,lgz=c
=(1/b+1/c)a+(1/a+1/c)b+(1/b+1/a)c
=a/b+a/c+b/a+b/c+c/b+c/a
=(a+c)/b+(a+b)/c+(b+c)/a
=-1-1-1
=-3
所以 lg[x^(1/b+1/c) ]+lg[y^(1/a+1/c) ]+lg[z^(1/b+1/a)]=-3
x^(1/b+1/c) y^(1/a+1/c) z^(1/b+1/a)=10^(-3)=1/1000