# 108中正預校國中部

## 回復 10# satsuki931000 的帖子

\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}

TOP

## 回復 10# satsuki931000 的帖子

\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}

TOP

## 回復 10# satsuki931000 的帖子

\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}

\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 ]

TOP

## 回復 10# satsuki931000 的帖子

K

#### 附件

85396E9F-23E9-41B8-9723-1880231FBEA0.jpeg (533.6 KB)

2019-5-7 14:50

TOP

## 回復 10# satsuki931000 的帖子

\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}

TOP

TOP

## 回復 16# q1214951 的帖子

#### 附件

1E2D9EEA-0C48-4E94-B899-EE04850529B2.jpeg (615.87 KB)

2019-5-7 19:52

TOP

TOP