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