ta có x_2=\frac{aX_1-1}{x_1+1}=(\frac{y-1}{y+1}-1):(\frac{y-1}{y+1}+1)=\frac{-2}{y+1}:\frac{2y}{y+1}=\frac{-2}{y+1}.\frac{y+1}{2y}=\frac{-1}{y}tương tự ta đcx_3=\frac{y+1}{1-y};x_4=y suy ra x_5=x_1;x_6=x_2;x_7=x_3;....vì 1986=4.496+2,nên x_{1986}=x_2=\frac{-1}{y}=3 \Rightarrow y =\frac{-1}{3}
ta có x_2=\frac{aX_1-1}{x_1+1}=(\frac{y-1}{y+1}-1):(\frac{y-1}{y+1}+1)=\frac{-2}{y+1}:\frac{2y}{y+1}=\frac{-2}{y+1}.\frac{y+1}{2y}$=\frac{-1}{y}$tương tự ta đcx_3=\frac{y+1}{1-y};x_4=y suy ra x_5=x_1;x_6=x_2;x_7=x_3;....vì 1986=4.496+2,nên x_{1986}=x_2=\frac{-1}{y}=3 \Rightarrow y =\frac{-1}{3}