﻿ Math Pro 數學補給站

-------------------------------------

(也感謝熱心夥伴提供題目的完整性與修正)
-------------------------------------

108.5.7版主補充
4.

(2005AMC12，https://artofproblemsolving.com/ ... Problems/Problem_23)

16.

(1)描繪出函數$$y=-2+\sqrt{-x^2+6x+7}$$的圖形。
(2)求定積分$$\displaystyle \int_{-1}^7 -2+\sqrt{-x^2+6x+7}dx$$。(請試以高中理科數學所教定積分與面積關係解之)

1.

https://math.pro/db/attachment.php?aid=4904&k=0098ebca43379c8ac2f3a04ec52bed8a&t=1657039389

https://math.pro/db/attachment.php?aid=4923&k=93573140443a8ccf36b269dd236855a3&t=1657039389

https://math.pro/db/attachment.php?aid=5032&k=2e8ee8b70cb39340e1fb0787f922b7f2&t=1657039389

11題

$$M=x^4-y^4$$有極值時
$$x+y=n$$

[解答]

\begin{align} & \frac{n+2}{n!+\left( n+1 \right)!+\left( n+2 \right)!} \\ & =\frac{n+2}{n!\left( 1+n+1+{{n}^{2}}+3n+2 \right)} \\ & =\frac{1}{n!\left( n+2 \right)} \\ & =\frac{n+2-1}{\left( n+2 \right)!} \\ & =\frac{1}{\left( n+1 \right)!}-\frac{1}{\left( n+2 \right)!} \\ \end{align}

https://math.pro/db/attachment.php?aid=4905&k=5d8fdaa5ea9646c8f9eac59efb2c946a&t=1657039389

[解答]

(1) 其中 3 座相通(用 3 條通道)，另 2 座也相通(用 1 條通道)，但此二系統不互通

(2) 其中 4 座相通(用 4 條通道)，另 1 座獨立

[解答]
\begin{align} & \alpha \beta =1 \\ & {{S}_{1}}=\alpha +\beta =1 \\ & {{S}_{2}}={{\alpha }^{2}}+{{\beta }^{2}}=-1 \\ & {{S}_{3}}=\left( \alpha +\beta \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-\alpha \beta \left( \alpha +\beta \right)={{S}_{2}}-{{S}_{1}}=-2 \\ & {{S}_{4}}=\left( \alpha +\beta \right)\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)-\alpha \beta \left( {{\alpha }^{2}}+{{\beta }^{2}} \right)={{S}_{3}}-{{S}_{2}}=-1 \\ & {{S}_{5}}={{S}_{4}}-{{S}_{3}}=1 \\ & {{S}_{6}}={{S}_{5}}-{{S}_{4}}=2 \\ & {{S}_{7}}={{S}_{6}}-{{S}_{5}}=1 \\ & {{S}_{8}}={{S}_{7}}-{{S}_{6}}=-1 \\ \end{align}

[ 本帖最後由 z78569 於 2019-4-21 13:14 編輯 ]

(在看有沒有其他夥伴能夠提供更正確的題目

[ 本帖最後由 z78569 於 2019-4-21 18:13 編輯 ]

https://math.pro/db/attachment.php?aid=4907&k=ddccfb2ca492791fe7d622238b767b80&t=1657039389

##### 引用:

[解答]
W.L.O.G 假設OA=OB=OC=OD=1

https://math.pro/db/attachment.php?aid=4909&k=66cfde919e85cba4984d764ae439dcac&t=1657039389

https://math.pro/db/attachment.php?aid=4912&k=5f6263a9ab8240748934a8c70654f11f&t=1657039389

$$x,y$$為實數且$$x>y$$，若$$x+y=x^2+y^2$$，則當$$x+y=n$$時，$$x^4-y^4$$有最大值$$M$$，求數對$$(n,M)=$$
[解答]
\begin{align} & {{x}^{2}}+{{y}^{2}}=x+y=n \\ & xy=\frac{{{\left( x+y \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}} \right)}{2}=\frac{{{n}^{2}}-n}{2} \\ & {{\left( x-y \right)}^{2}}={{\left( x+y \right)}^{2}}-4xy={{n}^{2}}-4\times \frac{{{n}^{2}}-n}{2}=-{{n}^{2}}+2n \\ & x-y=\pm \sqrt{-{{n}^{2}}+2n} \\ & {{x}^{4}}-{{y}^{4}}=\left( {{x}^{2}}+{{y}^{2}} \right)\left( x+y \right)\left( x-y \right)=\pm {{n}^{2}}\sqrt{-{{n}^{2}}+2n}=\pm \sqrt{-{{n}^{6}}+2{{n}^{5}}} \\ \end{align}

https://math.pro/db/attachment.php?aid=4915&k=41ba807b4c3f73476cb576e239b5ecd7&t=1657039389

https://math.pro/db/attachment.php?aid=4920&k=287c47358b37d6a0bc126785cd946e14&t=1657039389

[ 本帖最後由 Sandy 於 2019-4-24 15:48 編輯 ]

https://math.pro/db/attachment.php?aid=4922&k=cbc14845d323c3c6765fc89f494d0826&t=1657039389

https://math.pro/db/attachment.php?aid=4924&k=19288c06a66788137956bddd9c67f38e&t=1657039389

https://math.pro/db/attachment.php?aid=4925&k=418546f0f408099143e23a1cfbfeb731&t=1657039389

100 中壢高中和鳳山高中考過

[ 本帖最後由 thepiano 於 2019-4-25 00:01 編輯 ]

\begin{align} & P\left( x,y,z \right) \\ & x+y+z=0 \\ & \\ & {{\overline{PA}}^{2}}+{{\overline{PB}}^{2}}+{{\overline{PC}}^{2}} \\ & =3{{\left( x-1 \right)}^{2}}+3{{y}^{2}}+2+{{\left( z-2 \right)}^{2}}+{{z}^{2}}+{{\left( z-4 \right)}^{2}} \\ & =3\left[ {{\left( x-1 \right)}^{2}}+{{y}^{2}}+{{\left( z-2 \right)}^{2}} \right]+10 \\ & \ge 3\times \frac{{{\left( x+y+z-3 \right)}^{2}}}{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}+10 \\ & =19 \\ \end{align}

14另解

https://math.pro/db/attachment.php?aid=4926&k=25327c574eb5dc760ecc885d7b0f4a7a&t=1657039389

https://math.pro/db/attachment.php?aid=4927&k=309701afb4c2a07fb50e4670a3187b6c&t=1657039389

-

(不確定能不能這樣微分求值，不過答案是對的...)

[ 本帖最後由 royan0837 於 2019-4-25 15:52 編輯 ]

##### 引用:

-

$$\displaystyle=3^{n+1}\left(1+3+3^2+\cdots+3^{n-1}\right)-3\left(1+3+3^2+\cdots+3^{n-1}\right)$$

$$\displaystyle=3\left(3^n-1\right)\left(1+3+3^2+\cdots+3^{n-1}\right)$$

$$\displaystyle=3\left(3^n-1\right)\left(\frac{1\cdot\left(3^n-1\right)}{3-1}\right)$$

$$\displaystyle=\frac{3}{2}\left(3^n-1\right)^2$$

[ 本帖最後由 lyingheart 於 2019-10-24 21:32 編輯 ]

https://math.pro/db/attachment.php?aid=5741&k=4978f0731196c53f72e9024d6bdd9677&t=1657039389

 歡迎光臨 Math Pro 數學補給站 (https://math.pro/db/) 論壇程式使用 Discuz! 6.1.0