第5題
設\(x,y\)均大於0，則聯立方程組\(\cases{\displaystyle \frac{4x+3y-6}{4x+3y}=\frac{2}{\sqrt{x}}\cr \frac{4x+3y+6}{4x+3y}=\frac{4}{\sqrt{y}}}\)之解\((x,y)\)為　　　。
[解答]
\(\displaystyle z=\frac{6}{4x+3y} \)
\(\displaystyle \frac{4}{x}=1-2z+z^2 \)
\(\displaystyle \frac{16}{y}=1+2z+z^2 \)
\(\displaystyle \frac{16}{y}-\frac{4}{x}=4z \)
\(\displaystyle \frac{4x-y}{xy}=\frac{6}{4x+3y} \)
\(\displaystyle 16x^2+8xy-3y^2=6xy \)
\(\displaystyle (8x-3y)(2x+y)=0 \)
\(\displaystyle 8x=3y \)
\(\displaystyle 1-\frac{6}{12x}=\frac{2}{\sqrt x} \)
\(\displaystyle 2x-4\sqrt x-1=0 \)
\(\displaystyle \sqrt x=\frac{2+\sqrt6}{2} \)
\(\displaystyle x=\frac{5+2\sqrt6}{2} \)
\(\displaystyle y=\frac{20+8\sqrt6}{3} \)


第8題
令\(\Delta ABC\)的外心為\(O\)，垂心為\(H\)。若\(\Delta ABC\)的外接圓半徑為3，\(∠A=60^{\circ}\)，\(∠B=45^{\circ}\)，則\(\overline{OH}^2\)之值為　　　。
[解答]
由尤拉線性質知道
\(\displaystyle \vec{OH}=3\vec{OG}=\vec{OA}+\vec{OB}+\vec{OC} \)
然後就平方計算

第9題
不等式\(log_{14}(\sqrt{x}+\root 3\of x+\root 6 \of x)>log_{64}x\)之解集合為　　　。
[解答]
令 \(\displaystyle \sqrt[6]{x}=k \)
原式整理成  
\(\displaystyle \log_{14}(k+k^2+k^3) > \log_2 k \)
再假設  \(\displaystyle \log_2 k=y, \Rightarrow k=2^y \)
又可整理成
\(\displaystyle \log_{14}(2^y+4^y+8^y) > y \)
\(\displaystyle 2^y+4^y+8^y > 14^y \)
\(\displaystyle (\frac{1}{7})^y+(\frac{2}{7})^y+(\frac{4}{7})^y > 1 \)

當 \( y=1 \) 時，\(\displaystyle \frac{1}{7}+\frac{2}{7}+\frac{4}{7} = 1 \)
所以當 \(\displaystyle y > 1 \) 時，
\(\displaystyle (\frac{1}{7})^y+(\frac{2}{7})^y+(\frac{4}{7})^y < \frac{1}{7}+\frac{2}{7}+\frac{4}{7}=1 \)
當 \(\displaystyle y < 1 \) 時，
\(\displaystyle (\frac{1}{7})^y+(\frac{2}{7})^y+(\frac{4}{7})^y > \frac{1}{7}+\frac{2}{7}+\frac{4}{7}=1 \)
所以解為 \(\displaystyle y < 1 \)
即 \(\displaystyle 0 < k < 2 \)
\(\displaystyle 0 < x < 64 \)