Question

已知x,y,z满足x+(1/y)=4,y+(1/z)=1,z+(1/x)=7/3,则xyz的值为？

Answer 1

你好

(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

Answer 2

解：
x+ 1/y=4 x=4 -1/y=(4y-1)/y
y+1/z=1 1/z=1-y z=1/(1-y)
z+(1/x)=7/3
1/(1-y) +y/(4y-1)=7/3
整理，得
3(4y-1)+3y(1-y)=7(1-y)(4y-1)
25y²-20y+4=0
(5y-2)²=0
y=2/5 x=4- 1/y=4- 5/2=3/2 z=1/(1-y)=1/(1- 2/5)=5/3
x=3/2 y=2/5 z=5/3
xyz=(3/2)(2/5)(5/3)=1