回復 1# ycdye 的帖子
第一題：
\(\displaystyle 1+\frac{1}{\left(a-1\right)\left(a+1\right)}=\frac{a^2}{\left(a-1\right)\left(a+1\right)}=\frac{a}{a-1}\cdot\frac{a}{a+1}\)
因此，\(\displaystyle k=\left(\frac{2}{1}\cdot\frac{2}{3}\right)\left(\frac{3}{2}\cdot\frac{3}{4}\right)\left(\frac{4}{3}\cdot\frac{4}{5}\right)\cdots\left(\frac{1001}{1000}\cdot\frac{1001}{1002}\right)\)
\(\displaystyle =\left(\frac{2}{1}\cdot\not\frac{2}{3}\right)\left(\not\frac{3}{2}\cdot\not\frac{3}{4}\right)\left(\not\frac{4}{3}\cdot\not\frac{4}{5}\right)\cdots\left(\not\frac{1001}{1000}\cdot\frac{1001}{1002}\right)\)
\(\displaystyle =\frac{2}{1}\cdot\frac{1001}{1002}=\frac{1001}{501}\)
\(\displaystyle \Rightarrow \frac{3}{2}<k\leq2\)
第二題：
\(\displaystyle \left(a^2+x^2\right)^2=\left(a^2-x^2\right)^2+4a^2x^2\)
由題目條件的 \(\displaystyle a^2-x^2=ax\) 帶入上式，
可得 \(\displaystyle \left(a^2+x^2\right)^2=\left(ax\right)^2+4a^2x^2\)
\(\displaystyle \Rightarrow \left(a^2+x^2\right)^2=5a^2x^2\)
又 \(a>0\) 且 \(x>0\)，可得
\(\displaystyle \Rightarrow a^2+x^2=\sqrt{5}ax\)
\(\displaystyle \Rightarrow \frac{a^2+x^2}{ax}=\sqrt{5}\)
\(\displaystyle \Rightarrow \frac{a^2+x^2}{2ax}=\frac{\sqrt{5}}{2}\)
另解：
\(a^2-x^2=ax\)
\(\Rightarrow a^2-ax-x^2=0\)
\(\displaystyle \Rightarrow \left(a-\frac{1+\sqrt{5}}{2}x\right)\left(a-\frac{1-\sqrt{5}}{2}x\right)=0\)
\(\displaystyle \Rightarrow a=\frac{1+\sqrt{5}}{2}x\) 或 \(\displaystyle a=\frac{1-\sqrt{5}}{2}x\)（不合，因為 \(a>0\) 且 \(x>0\)）
剩下部分（將上式的 \(a\) 帶入題目所求式子）略。
