老師們好，
想請教台南一中科學班98複試跟103初試共6題，
煩請各位老師指教，謝謝。

6.角GCP=角GBP=角GBA
連BG交AC於K，令AK=X、PK=Y
由內分比得AB:AX=BP: PX，可得BP=2Y
PB+PC=2Y+(X-Y)=X+Y=PA

"一個圓的兩條弦"那題，題目有問題，BE不唯一

2
\(\begin{align}
  & {{T}_{n}}=\frac{2}{3}\times \frac{4}{3}\times \frac{4}{5}\times \frac{6}{5}\times \cdots \cdots \times \frac{2n}{2n+1}\times \frac{2n+2}{2n+1} \\
& {{T}_{n}}^{2}>\frac{1}{2}\times \frac{3}{2}\times \frac{2}{3}\times \frac{4}{3}\times \frac{3}{4}\times \frac{5}{4}\times \cdots \cdots \times \frac{2n-1}{2n}\times \frac{2n+1}{2n}\times \frac{2n}{2n+1}\times \frac{2n+2}{2n+1}=\frac{n+1}{2n+1} \\
& {{T}_{n}}^{2}<\frac{2}{3}\times \frac{4}{3}\times \frac{3}{4}\times \frac{5}{4}\times \frac{4}{5}\times \frac{6}{5}\times \cdots \cdots \times \frac{2n}{2n+1}\times \frac{2n+2}{2n+1}\times \frac{2n+1}{2n+2}\times \frac{2n+3}{2n+2}=\frac{2n+3}{3n+3} \\
\end{align}\)

2
\(\begin{align}
  & {{x}^{2}}+2xy-1=0 \\
& 2{{y}^{2}}+xy-2=0 \\
& 2\left( {{x}^{2}}+2xy \right)=2{{y}^{2}}+xy=2 \\
& 2{{x}^{2}}+3xy-2{{y}^{2}}=0 \\
& \left( x+2y \right)\left( 2x-y \right)=0 \\
& y=2x \\
& ...... \\
&  \\
\end{align}\)

2
\(\begin{align}
  & {{x}^{2}}+xy+{{y}^{2}}-1=0 \\
& {{y}^{2}}-4\left( {{y}^{2}}-1 \right)\ge 0 \\
& ...... \\
\end{align}\)

15
\(\begin{align}
  & x\left( {{x}^{2}}-yz \right)=3 \\
& y\left( {{y}^{2}}-xz \right)=9 \\
& z\left( {{z}^{2}}-xy \right)=30 \\
&  \\
& xyz={{x}^{3}}-3={{y}^{3}}-9={{z}^{3}}-30 \\
& y=\sqrt[3]{{{x}^{3}}+6},z=\sqrt[3]{{{x}^{3}}+27} \\
&  \\
& x\sqrt[3]{{{x}^{3}}+6}\sqrt[3]{{{x}^{3}}+27}={{x}^{3}}-3 \\
& {{x}^{3}}\left( {{x}^{3}}+6 \right)\left( {{x}^{3}}+27 \right)={{\left( {{x}^{3}}-3 \right)}^{3}} \\
& 42{{x}^{6}}+135{{x}^{3}}+27=0 \\
& \left( {{x}^{3}}+3 \right)\left( 42{{x}^{3}}+9 \right)=0 \\
& {{x}^{3}}=-3\ or\ -\frac{3}{14} \\
& xyz=-6\ or\ -\frac{45}{14} \\
& \left( x,y,z \right)=\left( -\sqrt[3]{3},\sqrt[3]{3},2\sqrt[3]{3} \right) \\
\end{align}\)

抱歉,第2題，第二個方程式抄錯，但是不影響最後的解答

謝謝老師們的提點，感激不盡~~

另想請教一下thepiano老師，
對於一個圓的兩條弦的那題，老師的看法為何？
有點不解為何線段BE不唯一，
題目我是直接從南一中考古題電子檔剪下來的，
所以應該是沒有打字出錯的疑慮，
再煩請老師指點，謝謝您。^^