﻿ 111桃園高中(頁 1) - 高中的數學 - Math Pro 數學補給站

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### 111桃園高中

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### 回復 1# Gary 的帖子

111.4.25補充

1/a^3(b+c) + 1/b^3(a+c) + 1/c^3(a+b) >= (1/a+1/b+1/c)/2

b^2c^2/a(b+c) + a^2c^2/b(a+c) + a^2b^2/c(a+b) >=(科西) (ab+bc+ab)^2/(2(ab+bc+ac)) =(ab+bc+ac)/2 = (1/a+1/b+1/c)/2

x=w, x=w^2 ,.., x=w^2022 帶入相加，可得 P(1)=0

[]表高斯符號，求解$$3x^2-19 \cdot [\;x ]\;+20=0$$。
[img]https://math.pro/db/attachment.php?aid=3457&k=0588e6cee6e335b9d13161147dd96bdb&t=1650773552&noupdate=yes[/img]

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[img]https://math.pro/db/attachment.php?aid=2543&k=f14a7befb9dd9621d47e405283b4607f&t=1650769246&noupdate=yes[/img]

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When $$4444^{4444}$$ is written in decimal notation, the sum of its digits is $$A$$. Let $$B$$ be the sum of the digits of $$A$$. Find the sum of the digits of $$B$$. ($$A$$ and $$B$$ are written in decimal notation.)
1975IMO，[url]https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_4[/url]

Let $$a, b, c$$ be positive real numbers such that $$abc=1$$. Prove that $$\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac{3}{2}.$$
1995IMO，[url]https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_2[/url] 填充4

[疑問]

1+√ (sinx)-√ x =cos(2x)+2x²  ,

(√ (sinx)-√ x) [2√ (sinx)+2√ x+1]=0
√ (sinx)-√ x=0 =>sinx=x-------(*)
∵0≦x≦π,畫圖可知僅當x=0時

[解答] 填充8

[解答] 填13.

[解答] 填13

[解答]

### 回復 14# Joanna 的帖子

[解答]

a_1 + a_2 + a_3 + a_4 + a_5 = 16 的正整數解

5 次連續下雨之間有 4 次未連續下雨

c_1 + c_2 + c_3 + c_4 + c_5 + c_6 = 12

#10

[解答]
z*w+(1/2)w*i=3i-(1/2)ŵ*i +z    (ŵ表示w bar)
z(w-1)=i [ 3-1/2(ŵ+w) ]    (令w=x+y*i)
3* √[(x-1)² +y²] = |3- x|

w所形成圖形為一個橢圓Γ,中心(3/4,0)

a=3/4 ,c=1/4,所以Γ的其中一個焦點為F1(3/4-1/4,0)=(1/2,0)

### 回復 17# ChuCH 的帖子

[解答]
$$\displaystyle \int_{1}^{\sqrt{3}}\sqrt{1+x^2}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\sec^3\theta d\theta=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\sec\theta d(\tan{\theta})$$
$$\displaystyle =\sec{\theta}\tan{\theta}-\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\sec{\theta}\tan^2{\theta}d\theta$$
$$\displaystyle =\sec{\theta}\tan{\theta}-\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\sec\theta(\sec^2\theta-1)d\theta$$
$$\displaystyle \Rightarrow\int\sec^3\theta d\theta =\frac{1}{2}(\sec\theta\tan\theta +\int sec\theta d\theta+C)$$
($$\displaystyle \int sec\theta d\theta=\int \sec\theta \frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}d\theta=\int \frac{1}{\sec\theta+\tan\theta} d(\sec\theta+\tan\theta)=\ln |\sec\theta+\tan\theta|+c$$)