诺实数x,y,z满足x+1/y=4,y+1/z=1,z+1/x=7/3,求xyz

Question

诺实数x,y,z满足x+1/y=4,y+1/z=1,z+1/x=7/3,求xyz

Answer 1

(X+1/Y)(Z+1/X)(Y+1/Z)=XYZ+1/(XYZ)+X+Y+Z+1/X+1/Y+1/Z=XYZ+1/(XYZ)+4+1+7/3=4*1*7/3
得到：
XYZ+1/(XYZ)=2,解得XYZ=1