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2018.07.06 筆試

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\begin{align} & {{\left( 1+\frac{1}{x} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{x} \right)}^{2}}={{\left( x+1 \right)}^{2}} \\ & \sec B=x=\sqrt[3]{2} \\ \end{align}

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[ 本帖最後由 laylay 於 2018-7-15 15:04 編輯 ]

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$$\displaystyle f\left(x\right)=\left|4x-3a\right|+\left|5x-4a\right|=4\left|x-\frac{3a}{4}\right|+5\left|x-\frac{4a}{5}\right|$$

$$\displaystyle =4\left(\left|x-\frac{3a}{4}\right|+\left|x-\frac{4a}{5}\right|\right)+\left|x-\frac{4a}{5}\right|$$

$$\displaystyle \geq 4\left|\left(x-\frac{3a}{4}\right)-\left(x-\frac{4a}{5}\right)\right|+\left|x-\frac{4a}{5}\right|$$

$$\displaystyle =\frac{\left|a\right|}{5}+\left|x-\frac{4a}{5}\right|$$

$$\displaystyle\frac{\left|a\right|}{5}\geq a^2\Rightarrow \frac{\left|a\right|}{5} \geq \left|a\right|^2\Rightarrow \frac{-1}{5}\leq a\leq \frac{1}{5}$$

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