若 $a,b,c,d,e$ 是不全相等的实数, 且 $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{d}=d+\frac{1}{e}=e+\frac{1}{a}=x$, 求 $x$. 且证明: $abcde+x=0$.
类似问题: 2170
1
根据题设,
\[
\begin{aligned}
b&=\frac{1}{x-a},\\
c&=\frac{1}{x-b},\\
d&=\frac{1}{x-c},\\
e&=\frac{1}{x-d},\\
a&=\frac{1}{x-e}.\\
\end{aligned}
\]
于是,
\[
\begin{split}
a&=\frac{1}{x-e}=\frac{1}{x-\frac{1}{x-d}}=\frac{1}{x-\frac{1}{x-\frac{1}{x-c}}}\\
&=\frac{1}{x-\frac{1}{x-\frac{1}{x-\frac{1}{x-b}}}}=\frac{1}{x-\frac{1}{x-\frac{1}{x-\frac{1}{x-\frac{1}{x-a}}}}}\\
&=\frac{1}{x-\frac{1}{x-\frac{1}{x-\frac{x-a}{x^2-ax-1}}}}=\frac{1}{x-\frac{1}{x-\frac{x^2-ax-1}{x^3-ax^2-2x+a}}}\\
&=\frac{1}{x-\frac{x^3-ax^2-2x+a}{x^4-ax^3-3x^2+2ax+1}}=\frac{x^4-ax^3-3x^2+2ax+1}{x^5-ax^4-4x^3+3ax^2+3x-a}
\end{split}
\]
这推出
\[
\begin{split}
&ax^5-a^2 x^4-4ax^3+3a^2 x^2+3ax-a^2=x^4-ax^3-3x^2+2ax+1\\
\Rightarrow\ &ax^5-(a^2+1)x^4-3ax^3+3(a^2+1)x^2+ax-(a^2+1)=0\\
\Rightarrow\ &(ax^5-3ax^3)-(a^2+1)(x^4-3x^2)+ax-(a^2+1)=0\\
\Rightarrow\ &ax^3(x^2-3)-(a^2+1)x^2(x^2-3)+\bigl[ax-(a^2+1)\bigr]=0\\
\Rightarrow\ &x^2(x^2-3)\bigl[ax-(a^2+1)\bigr]+\bigl[ax-(a^2+1)\bigr]=0\\
\Rightarrow\ &(x^4-3x^2+1)\cdot\bigl[ax-(a^2+1)\bigr]=0\\
\end{split}
\]
因此 $x^4-3x^2+1=0$ 或 $ax=a^2+1$.
由 $x^4-3x^2+1=0$ 推出
\[
x^2=\frac{3\pm\sqrt{5}}{2}
\]
而后式推出 $x=a+\frac{1}{a}$, 注意 $a\neq 0$.
若 $x=a+\frac{1}{a}$, 则 $b=\frac{1}{x-a}=a$, $c=\frac{1}{x-b}=\frac{1}{x-a}=b$, $d=\frac{1}{x-c}=\frac{1}{x-b}=c$, $e=\frac{1}{x-d}=\frac{1}{x-c}=d$, $a=\frac{1}{x-e}=\frac{1}{x-d}=e$. 即 $a=b=c=d=e$ 与题设矛盾.
因此,
\[
x^2=\frac{3\pm\sqrt{5}}{2}=\frac{6\pm 2\sqrt{5}}{4}=(\frac{\sqrt{5}\pm 1}{2})^2
\]
即
\[
x=\pm\frac{\sqrt{5}+1}{2},\quad{\text{或}}x=\pm\frac{\sqrt{5}-1}{2}.
\]