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98嘉義高工

98嘉義高工

我是第一次在這發文,若有不禮貌的發文,還請告知,我想請問以下題目,
(1)若a,b,c為相異正實數,則(a)^(logb/c)*(b)^(logc/a)*(c)^(loga/b)=?
(2)|向量a|=|向量b|不等於0,若|向量a+向量b|-|向量a-向量b|=根號|向量a|,則向量a和向量b的夾角為何?
(3)設x,y,z為自然數,x<y<z,且其中任二數之和均為另一數的倍數,求 x:y:z=?
謝謝
(因為我不會用數學符號,請問該如何用數學符號)
(請問版主可否用word 打好,再用附上呢?)

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附上題目,以後寫題號就可以了

附件

98嘉義高工.rar (129.71 KB)

2010-8-25 06:35, 下載次數: 5799

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第 4 題

題目:若 \(\displaystyle a,b,c\) 為相異正實數,則 \(\displaystyle a^{log\frac{b}{c}}\cdot b^{log\frac{c}{a}} \cdot c^{log\frac{a}{b}}=\)______________。

解答:

\(\displaystyle \log\left(a^{log\frac{b}{c}}\cdot b^{log\frac{c}{a}} \cdot c^{log\frac{a}{b}}\right)\)

\(\displaystyle =log\frac{b}{c}\cdot\log a + log\frac{c}{a}\cdot \log b + log\frac{a}{b}\cdot \log c\)

\(\displaystyle =\left(\log b - \log c\right)\log a +\left(\log c - \log a\right)\log b +\left(\log a - \log b\right)\log c\)

\(\displaystyle =0\)

故,\(\displaystyle a^{log\frac{b}{c}}\cdot b^{log\frac{c}{a}} \cdot c^{log\frac{a}{b}} = 1\)




第 6 題

題目: \(\displaystyle \left|\vec{a}\right|=\left|\vec{b}\right|\neq0\),若 \(\displaystyle \left|\vec{a}+\vec{b}\right|-\left|\vec{a}-\vec{b}\right|=\sqrt{2}\left|\vec{a}\right|\),則 \(\displaystyle \vec{a}\) 與 \(\displaystyle \vec{b}\) 的夾角為____________。

解答:

令 \(\displaystyle \vec{a}\) 與 \(\displaystyle \vec{b}\) 的夾角為 \(\displaystyle \theta\),\(\displaystyle \left|\vec{a}\right|=k>0\),



\(\displaystyle \left|\vec{a}+\vec{b}\right|=\sqrt{\left(\vec{a}+\vec{b}\right)\cdot\left(\vec{a}+\vec{b}\right)}=k\sqrt{2\left(1+\cos\theta\right)}\)

\(\displaystyle \left|\vec{a}-\vec{b}\right|=\sqrt{\left(\vec{a}-\vec{b}\right)\cdot\left(\vec{a}-\vec{b}\right)}=k\sqrt{2\left(1-\cos\theta\right)}\)

因此,可得 \(\displaystyle \sqrt{2\left(1+\cos\theta\right)} - \sqrt{2\left(1-\cos\theta\right)} = \sqrt{2}\)

\(\displaystyle \Rightarrow \sqrt{1+\cos\theta} - \sqrt{1-\cos\theta} = 1\)

\(\displaystyle \Rightarrow \sqrt{1+\left(2\cos^2\frac{\theta}{2}-1\right)} - \sqrt{1-\left(1-2\sin^2\frac{\theta}{2}\right)} = 1\)

\(\displaystyle \Rightarrow \left|\cos\frac{\theta}{2}\right|-\left|\sin\frac{\theta}{2}\right|=\frac{1}{\sqrt{2}}\)

因為 \(0^\circ\leq\theta\leq180^\circ\),所以 \(\displaystyle \cos\frac{\theta}{2}-\sin\frac{\theta}{2}=\frac{1}{\sqrt{2}}\)

\(\displaystyle \Rightarrow \sqrt{2}\cos\left(\frac{\theta}{2}+45^\circ\right)=\frac{1}{\sqrt{2}}\)

\(\displaystyle \Rightarrow \cos\left(\frac{\theta}{2}+45^\circ\right)=\frac{1}{2}\)

\(\displaystyle \Rightarrow \frac{\theta}{2}+45^\circ=\pm 60^\circ + 360^\circ\cdot k,\,k\in\mathbb{Z}\)

因為 \(0^\circ\leq\theta\leq180^\circ\),所以 \(\theta=30^\circ.\)

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第 8 題

題目:設 \(\displaystyle x,y,z\) 為自然數,\(\displaystyle x<y<z\),且其中任二數之和均為另一數的倍數,求 \(\displaystyle x:y:z=\)_______。


解答:

因為 \(\displaystyle 0<\frac{x+y}{z}<\frac{z+z}{z}=2\),所以 \(\displaystyle \frac{x+y}{z}=1\Rightarrow x+y=z\)


因為 \(\displaystyle \frac{x+z}{y}=\frac{x+\left(x+y\right)}{y}=\frac{2x+y}{y}\),且 \(\displaystyle 1=\frac{y}{y}<\frac{2x+y}{y}<\frac{3y}{y}=3\),


所以 \(\displaystyle \frac{x+z}{y}=2\Rightarrow x+z=2y\)




由 \(\displaystyle \left\{\begin{array}{ccc}x+y&=z\\x+z&=2y\end{array}\right.\Rightarrow\left\{\begin{array}{ccc}x+y-z&=0\\x-2y+z&=0\end{array}\right.\)

\(\Rightarrow x:y:z=1:2:3.\)

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謝謝

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想請問一下第7題,和第12題(最後一項分母突然出現2n,就不知道該怎麼算了)
謝謝

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回復 6# pizza 的帖子

第 7 題:

由柯西不等式,

可得 \(\displaystyle\left(x^2+\left(2y\right)^2+z^2\right)\left(2^2+\left(\frac{-1}{2}\right)^2+1^2\right)\geq\left(2x-y+z\right)^2\)

\(\displaystyle\Rightarrow \frac{21}{4}\geq\frac{\left(2x-y+z\right)^2}{x^2+4y^2+z^2}\)

\(\displaystyle\Rightarrow \frac{-\sqrt{21}}{2}\leq\frac{2x-y+z}{\sqrt{x^2+4y^2+z^2}}\leq\frac{\sqrt{21}}{2}.\)


第 12 題:

因為 \(\displaystyle\frac{1}{\sqrt{3n^2+2n}}\cdot 2n\leq a_n\leq\frac{1}{\sqrt{3n^2+1}}\cdot 2n\)

且 \(\displaystyle \lim_{n\to\infty}\frac{2n}{\sqrt{3n^2+2n}} =\lim_{n\to\infty}\frac{2}{\sqrt{3+\frac{2}{n}}}  =\frac{2}{\sqrt{3}} \)
且 \(\displaystyle
\lim_{n\to\infty}\frac{2n}{\sqrt{3n^2+1}} =\lim_{n\to\infty}\frac{2}{\sqrt{3+\frac{1}{n^2}}}

=\frac{2}{\sqrt{3}}\)

所以 \(\displaystyle \lim_{n\to\infty}a_n=\frac{2}{\sqrt{3}}.\)

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請教 填充11題,該如何做~ = =+

\( x,y \in R \),\( x^2+y^2=1 \),若\( x^2+4xy+5y^2 \)的最大值\( =M \),最小值\( =m \),則\( M+m= \)   

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回復 8# mathelimit 的帖子

第 11 題
常見的考古題,請參考
https://math.pro/db/thread-882-1-1.html

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回復 9# thepiano 的帖子

算出來了~ 謝謝 ^^

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