题目

八下一道简单的数学题急 急
已知x+1y=y+1z=1
(1)求z+1x的值;
(2)如果实数x.y.z满足x+1y=4,y+1z=1,z+1x=73,求xyz的值.
(只知道一问就答一问)
急 急

提问时间：2020-10-22

答案

(1)x=1-1/y,1/x=y/(y-1) 1/z=1-y,z=1/(1-y)=-1/(y-1) z+1/x=y/(y-1)-1/(y-1)=1 (2)(x+1/y)(y+1/z)(z+1/x)=28/3(x+1/y)+(y+1/z)+(z+1/x)=22/7(x+1/y)(y+1/z)(z+1/x)=xyz+1/(xyz)+(x+1/y)+(y+1/z)+(z+1/x)=28/3xyz+1/(...