设y^2+4y+x=t
y^2+4y=t-x 代入方程:
x/(x-1)=(t-x-2)/(t-x-1)
tx-x^2-x=tx-x^2-2x-t+x+2
(t-1)x-x^2=-x^2+(t-2+1)x+2-t
(t-1-t+2-1)x=2-t
0=2-t
t=2
即:y^2+4y+x=2