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1.$$n=10$$
2.$$\displaystyle \frac{a_{3}}{a_{1}a_{2}}=50$$
3.覺得怪，有請高手講解...

1.n應該是10

2.考卷上的a_2是40，算法沒錯
3.我也覺得怪

n 是正整數嗎？

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https://math.pro/db/attachment.php?aid=3011&k=d5b6008c0a1ec6537c7ca581d043a36e&t=1660045589

$$a=8$$的時候  的解 是$$0<x<2$$ 這樣不滿足題意嗎??

$$a,b,c$$為三正數，且滿足$$abc(b+c)=5$$，則$$ab+bc+ca$$之最小值。

\begin{align} & ab+bc+ca \\ & =a\left( b+c \right)+bc \\ & =\frac{5}{bc}+bc \\ & \ge 2\sqrt{5} \\ \end{align}

$$\displaystyle a=\frac{tan \theta-sec \theta+1}{tan \theta+sec \theta-1}$$，若$$a$$是$$x^4-3x^3+2x^2-3x+1=0$$的解，求$$sin \theta=$$

https://math.pro/db/attachment.php?aid=3013&k=667569e6fa41167a0ac293869da34e01&t=1660045589

$$\left( \int_{1}^{5}{{{\left( \sqrt{2{{x}^{2}}-1} \right)}^{2}}dx-\int_{1}^{5}{{{\left( \frac{3}{2}x-\frac{1}{2} \right)}^{2}}dx}} \right)\pi$$

\begin{align} & {{c}^{2}}={{a}^{2}}+{{b}^{2}}\ge 2ab \\ & c\ge \sqrt{2}\sqrt{ab} \\ \end{align}

\begin{align} & a+b+c\ge \left( 2+\sqrt{2} \right)\sqrt{ab} \\ & \sqrt{ab}\le \frac{10}{2+\sqrt{2}} \\ & \frac{ab}{2}\le 25\left( 3-2\sqrt{2} \right) \\ \end{align}

(錯了，樓下鋼琴大有反例)負的應該"不合"

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$$\displaystyle \frac{tan\theta-sec\theta+1}{tan\theta+sec\theta-1}\times \frac{cos\theta}{cos\theta}$$
$$\displaystyle =\frac{sin\theta-1+cos\theta}{sin\theta+1-cos\theta}$$
$$\displaystyle =\frac{sin\theta-(1-cos\theta)}{sin\theta+(1-cos\theta)}$$

\begin{align} & \theta =-\frac{\pi }{4} \\ & \frac{\sin \theta -\left( 1-\cos \theta \right)}{\sin \theta +\left( 1-\cos \theta \right)}=\sqrt{2}+1>1 \\ \end{align}

https://math.pro/db/attachment.php?aid=3016&k=7627edb18aedc57431a5649f13faa541&t=1660045589

16

$$\displaystyle \int_{1}^{5}[\sqrt{2x^{2}-1}-(\frac{3}{2}x-\frac{1}{2})]dx$$

[解答]
AB' = AB = 8
AC' = AC = 6
∠C'AB' = 3∠BAC = 120 度

[解答]
\begin{align} & {{a}_{{{2}^{2015}}+1}} \\ & =2{{a}_{{{2}^{2014}}}}+1 \\ & =2\left( 2{{a}_{{{2}^{2013}}}}-1 \right)+1 \\ & ={{2}^{2}}{{a}_{{{2}^{2013}}}}-2+1 \\ & ={{2}^{3}}{{a}_{{{2}^{2012}}}}-{{2}^{2}}-2+1 \\ & ={{2}^{2015}}{{a}_{1}}-{{2}^{2014}}-{{2}^{2013}}-\cdots \cdots -{{2}^{2}}-2+1 \\ & =3 \\ \end{align}

TRML 2001 個人賽第 2 題

[解答]

111.1.27新增圖片

https://math.pro/db/attachment.php?aid=6187&k=c6c4d2286ff5b96d68b045b8d48e8d1e&t=1660045589

https://math.pro/db/attachment.php?aid=6188&k=dcf9a69215a70ce899680b2902f4f14a&t=1660045589

[解答]

[解答]

https://math.pro/db/attachment.php?aid=6150&k=c27c788f0927c9d7abe670bda0af3b44&t=1660045589

(1) 三場就勝

(2) 到第四場才勝

(3) 到第五場才勝

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