選擇第6題
作\(\overline{AM}\)垂直\(\overline{PQ}\)於\(M\)；作\(\overline{RN}\)垂直\(\overline{AM}\)於\(N\)
\(\begin{align}
  & \overline{AR}=2,\overline{MR}=\frac{1}{\sqrt{2}},\overline{AM}=\frac{3}{\sqrt{2}} \\
& \overline{AN}=x,\overline{MN}=\frac{3}{\sqrt{2}}-x \\
& \overline{RN}={{2}^{2}}-{{x}^{2}}={{\left( \frac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \frac{3}{\sqrt{2}}-x \right)}^{2}} \\
& x=\frac{4}{3}\sqrt{2} \\
& \overline{RN}=\frac{2}{3} \\
\end{align}\)

填充第D題
\(\begin{align}
  & f\left( x+2 \right)=\frac{1+f\left( x \right)}{1-f\left( x \right)} \\
& f\left( x+4 \right)=\frac{1+f\left( x+2 \right)}{1-f\left( x+2 \right)}=\frac{1+\frac{1+f\left( x \right)}{1-f\left( x \right)}}{1-\frac{1+f\left( x \right)}{1-f\left( x \right)}}=\frac{1}{-f\left( x \right)} \\
& f\left( x+6 \right)=\frac{1+f\left( x+4 \right)}{1-f\left( x+4 \right)}=\frac{1+\frac{1}{-f\left( x \right)}}{1-\frac{1}{-f\left( x \right)}}=\frac{f\left( x \right)-1}{f\left( x \right)+1} \\
& f\left( x+8 \right)=\frac{1+f\left( x+6 \right)}{1-f\left( x+6 \right)}=\frac{1+\frac{f\left( x \right)-1}{f\left( x \right)+1}}{1-\frac{f\left( x \right)-1}{f\left( x \right)+1}}=f\left( x \right) \\
\end{align}\)
剩下的就簡單了

填充第E題
\(\begin{align}
  & \sin x+\cos x=t \\
& -\sqrt{2}\le t\le \sqrt{2} \\
& \left| \sin x+\cos x+\tan x+\cot x+\sec x+\csc x \right| \\
& =\left| \sin x+\cos x+\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x} \right| \\
& =\left| \sin x+\cos x+\frac{1+\sin x+\cos x}{\sin x\cos x} \right| \\
& =\left| t+\frac{1+t}{\frac{{{t}^{2}}-1}{2}} \right| \\
& =\left| t+\frac{2}{t-1} \right| \\
\end{align}\)

微分可知
\(\begin{align}
  & t+\frac{2}{t-1}\ge 1+2\sqrt{2}\ or\ t\le 1-2\sqrt{2} \\
& \left| t+\frac{2}{t-1} \right|\ge 2\sqrt{2}-1 \\
\end{align}\)

[ 本帖最後由 thepiano 於 2019-5-23 15:31 編輯 ]

K
把正四面體展開
圖片中，紅色直線長即為所求。
可以自己摺一個，比較具體。

選擇第8題
\(\begin{align}
  & {{M}_{900}}=1 \\
&  \\
& 0={{a}_{1}}={{a}_{2}}=\cdots ={{a}_{898}}\le {{a}_{899}}\le {{a}_{900}} \\
& 1={{a}_{899}}^{2}+{{a}_{900}}^{2}\ge 2{{a}_{899}}^{2} \\
& {{a}_{899}}\le \frac{1}{\sqrt{2}},{{M}_{899}}=\frac{1}{\sqrt{2}} \\
&  \\
& \sum\limits_{n=1}^{900}{{{M}_{n}}}=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{900}} \\
\end{align}\)
剩下的就老梗題了

想請教填充A，謝謝老師！

填充A應該是這樣~
不過怕有地方寫錯

謝謝 roger0315 老師！

附圖為我的想法，請參考指教！

想請教填充M跟P，謝謝老師！