题目

解方程组(xy+x+y+1)/(x+y+2)=2
(xz+x+z+1)/(x+z+2)=3
(y+1)(z+1)/y+z+2=4
我希望能是具体步骤``谢谢`

提问时间：2021-01-02

答案

(xy+x+y+1)/(x+y+2)=2 即(x+y+2)/(xy+x+y+1)=[(x+1)+(y+1)]/(x+1)(y+1)
=1/(x+1)+1/(y+1)=1/2
同理：1/(x+1)+1/(z+1)=1/3 1/(y+1)+1/(z+1)=1/4
所以：三式一加：2[1/(x+1)+1/(y+1))+1/(z+1)]=1/2+1/3+1/4=13/12
1/(x+1)+1/(y+1)+1/(z+1)=13/24
所以1/(z+1)＝13/24-1/2＝1/24,z＝23
1/(y+1)＝13/24-1/3＝5/24,y＝24/5-1＝19/5
1/(x+1)＝13/24-1/4＝7/24,x＝24/7-1＝17/7