已知互不相等的四个实数\(a,b,c,d\)满足\[a+\dfrac 1b=b+\dfrac 1c=c+\dfrac 1d=d+\dfrac 1a=x,\]求\(x\)的所有可能的值．
正确答案是\(\pm\sqrt 2\)．
依次消元，根据条件有\[\begin{split}d&=x-\dfrac 1a\\c&=x-\dfrac 1d=\dfrac{ax^2-x-a}{ax-1}\\b&=x-\dfrac 1c=\dfrac{ax^3-x^2-2ax+1}{ax^2-x-a}\\a&=x-\dfrac 1b=\dfrac{ax^4-x^3-3ax^2+2x+a}{ax^3-x^2-2ax+1}\end{split}\]整理得\[x\left(x^2-2\right)\left(a^2-xa+1\right)=0,\]于是\[x=0\lor x=\pm\sqrt 2\lor x=a+\dfrac 1a.\]经验证，只有\(x=\pm\sqrt 2\)符合题意．
例如：当\((a,b,c,d)=(1,\sqrt 2+1,-1,\sqrt 2-1)\)时，\(x=\sqrt 2\)．
当\((a,b,c,d)=(-1,-\sqrt 2-1,1,-\sqrt 2+1)\)时，\(x=-\sqrt 2\)．