填充第1題
\(\begin{align}
  & x=\frac{4y-3}{5} \\
& y>0,y\equiv 2\ or\ 7\ \left( \bmod \ 10 \right) \\
& y<0,-y\equiv 3\ or\ 8\ \left( \bmod \ 10 \right) \\
\end{align}\)
利用\({{60}^{2}}+{{80}^{2}}={{100}^{2}}\)
\(\left( x,y \right)=\left( -59,-73 \right),\left( -55,-68 \right),\cdots ,\left( 57,72 \right),\left( 61,77 \right)\)

填充第3題

\(\begin{align}
  & {{a}_{n+1}}={{S}_{n}}+{{n}^{2}}-n+2 \\
& {{a}_{n}}={{S}_{n-1}}+{{\left( n-1 \right)}^{2}}-\left( n-1 \right)+2 \\
& {{a}_{n+1}}-a{}_{n}={{a}_{n}}+2n-2 \\
& {{a}_{n+1}}=2{{a}_{n}}+2\left( n-1 \right) \\
& {{a}_{n+1}}+2\left( n+1 \right)=2\left( {{a}_{n}}+2n \right)={{2}^{2}}\left[ {{a}_{n-1}}+2\left( n-1 \right) \right]=\cdots ={{2}^{n}}\left( {{a}_{1}}+2 \right) \\
& {{a}_{n}}={{2}^{n+1}}-2n \\
\end{align}\)

計算第2題
即證明\(2\left( 1+r+{{r}^{2}}+{{r}^{3}}+{{r}^{4}} \right)\le 5\left( 1+{{r}^{4}} \right)\)

\(\begin{align}
  & 5\left( 1+{{r}^{4}} \right)-2\left( 1+r+{{r}^{2}}+{{r}^{3}}+{{r}^{4}} \right) \\
& =3{{r}^{4}}-2{{r}^{3}}-2{{r}^{2}}-2r+3 \\
& ={{\left( r-1 \right)}^{2}}\left( 3{{r}^{2}}+4r+3 \right)\ge 0 \\
&  \\
& 2\left( 1+r+{{r}^{2}}+{{r}^{3}}+{{r}^{4}} \right)\le 5\left( 1+{{r}^{4}} \right) \\
\end{align}\)

[ 本帖最後由 thepiano 於 2016-6-2 04:28 PM 編輯 ]

填充第2題
\(\begin{align}
  & {{c}_{ij}}=\sum\limits_{k=1}^{8}{\left( {{a}_{ik}}\times {{b}_{kj}} \right)}\quad \left( i=1\tilde{\ }6,j=1\tilde{\ }7 \right) \\
& =\sum\limits_{k=1}^{8}{\left[ \left( -i+k \right)\left( k-2j \right) \right]} \\
& =\sum\limits_{k=1}^{8}{\left[ 2ij-\left( i+2j \right)k+{{k}^{2}} \right]} \\
& =16ij-36\left( i+2j \right)+204 \\
& =\left( 4i-18 \right)\left( 4j-9 \right)+42 \\
\end{align}\)
易知\(i=6,j=7\)時，\({{c}_{ij}}\)有最大值；\(i=1,j=7\)時，\({{c}_{ij}}\)有最小值

計算第三題
作\(\overline{QE}\)垂直\(\overline{AP}\)於\(E\)，\(\overline{QF}\)垂直\(\overline{BC}\)於\(F\)
令\(\overline{QF}=x\)，則\(\overline{QE}=1-x\)
\(\begin{align}
  & \frac{\overline{AP}}{\overline{BC}}=\frac{\overline{QE}}{\overline{QF}} \\
& \overline{AP}=\frac{1-x}{x} \\
& \Delta APQ+\Delta QBC=\frac{1}{2}\times \frac{1-x}{x}\times \left( 1-x \right)+\frac{1}{2}\times 1\times x \\
& =\frac{{{x}^{2}}-2x+1}{2x}+\frac{x}{2} \\
& =x+\frac{1}{2x}-1 \\
& \ge \sqrt{2}-1 \\
\end{align}\)