解方程组：（xy+x）\(x+y+1)=2,（xz+2x）\(z+x+2)=3,(y+1)(z+2)\(y+z+3)=4

Question

解方程组：（xy+x）\(x+y+1)=2,（xz+2x）\(z+x+2)=3,(y+1)(z+2)\(y+z+3)=4

Answer 1

(XY+X)/(X+Y+1)=2
(x+y+1)/(xy+x)=1/2
(x+y+1)/x(y+1)=1/2
1/(y+1)+1/x=1/2
同样得：
1/(z+2)+1/x=1/3
1/(y+1)+1/(z+2)=1/4
解方程组得：
1/x=7/24,1/(y+1)=5/24,1/(z+2)=1/24
所以
x=24/7,y=19/5,z=22
x+y+z=29又8/35