设1995x3=1996y3=1997z3,xyz＞0,且有3√1995x3+1996y3+1997z3=3√1995+3√1996+3√1997.求1/x+1/y+1/z

Question

设1995x3=1996y3=1997z3,xyz＞0,且有3√1995 x3+1996y3+1997z3= 3√1995+3√1996+ 3√1997.求1/x+1/y+1/z

Answer 1

解如下,希望能帮到你,望采纳哈.
1995x^3=1996y^3=1997z^3=k^3
1995x^2=k^3/x,1996y^2=k^3/y,1997z^2=k^3/z
(1995x^2+1996y^2+1997z^2)的立方根
=(k^3/x+k^3/y+k^3/z)的立方根
1995x^3=1996y^3=1997z^3=k^3
1995=k^3/x^3,1996=k^3/y^3,1997=k^3/z^3
1995的立方根+1996的立方根+1997的立方根
=k/x+k/y+k/z
(1995x^2+1996y^2+1997z^2)的立方根=1995的立方根+1996的立方根+1997的立方根
(k^3/x+k^3/y+k^3/z)的立方根=k/x+k/y+k/z
(1/x+1/y+1/z)的立方根=1/x+1/y+1/z
1/x+1/y+1/z=(1/x+1/y+1/z)^3
(1/x+1/y+1/z)[(1/x+1/y+1/z)^2-1]=0
因为xyz>0
所以1/x+1/y+1/z=1