題目如下
不簡單喔

令\(f\left( 1 \right)=k>-1,k\ne 0\)
\(\begin{align}
  & f\left( 1 \right)\times f\left( f\left( 1 \right)+1 \right)=1 \\
& k\times f\left( k+1 \right)=1 \\
& f\left( k+1 \right)=\frac{1}{k} \\
&  \\
& f\left( k+1 \right)\times f\left( f\left( k+1 \right)+\frac{1}{k+1} \right)=1 \\
& f\left( \frac{1}{k}+\frac{1}{k+1} \right)=\frac{1}{f\left( k+1 \right)}=k=f\left( 1 \right) \\
& \frac{1}{k}+\frac{1}{k+1}=1 \\
& k=\frac{1\pm \sqrt{5}}{2} \\
\end{align}\)
\(k=\frac{1+\sqrt{5}}{2}\)不合，為什麼不合，就留給您了

移項，\(f\left( f\left( x \right)+\frac{1}{x} \right)=\frac{1}{f\left( x \right)}\), 故
        \(f\left( f\left( x \right)+\frac{1}{x} \right)f\left( f\left( f\left( x \right)+\frac{1}{x} \right)+\frac{1}{f\left( x \right)+\frac{1}{x}} \right)=\frac{1}{f\left( x \right)}\cdot f\left( \frac{1}{f\left( x \right)}+\frac{1}{f\left( x \right)+\frac{1}{x}} \right)=1\)
故
        \(f\left( \frac{1}{f\left( x \right)}+\frac{1}{f\left( x \right)+\frac{1}{x}} \right)=f\left( x \right)\)

因為\(f\)在\({{\mathbb{R}}^{+}}\)為嚴格遞增，有1對1的性質，故\(\frac{1}{f\left( x \right)}+\frac{1}{f\left( x \right)+\frac{1}{x}}=x\)
, 整理成\(x{{\left( f\left( x \right) \right)}^{2}}-f\left( x \right)-\frac{1}{x}=0\Rightarrow f\left( x \right)=\frac{1\pm \sqrt{5}}{2x}\)
(取正號時不合)，故\(f\left( 1 \right)=\frac{1-\sqrt{5}}{2}\)

[ 本帖最後由 hua0127 於 2014-7-1 09:58 PM 編輯 ]

wow!
看來我晚了一步
剛才自己才想出來的說~~~
不過這題也沒表面得嚇人
帶入兩次再配合條件"嚴格遞增"就結束了