若实数x,y,z满足x+1⼀y=1,y+1⼀z=4,z+1⼀x=7⼀3,则xyz=?

2025-05-22 14:04:58
推荐回答(1个)
回答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
看完了好评我哦~~