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# System prepended metadata

title: 挑戰題 7.
tags: [數與式]

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###### tags: `數與式`
# 挑戰題 7.

假設 $2021x^3=2022y^3=2023z^3$，其中 $xyz>0$，且 $\sqrt[3]{2021x^2+2022y^2+2023z^2}=\sqrt[3]{2021}+\sqrt[3]{2022}+\sqrt[3]{2023}$，試求 $\begin{align}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\underline{\qquad\qquad}\end{align}$

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$\boxed{答}：1$
$\boxed{解}：$ 令 $2021x^3=2022y^3=2023z^3=t$
$\begin{align}\Rightarrow \sqrt[3]{\frac{t}{x}+\frac{t}{y}+\frac{t}{z}}=\sqrt[3]{\frac{t}{x^3}}+\sqrt[3]{\frac{t}{y^3}}+\sqrt[3]{\frac{t}{z^3}}\end{align}$
$\begin{align}\Rightarrow\sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\end{align}$
$\begin{align}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0，1，-1\end{align}$
又因為 $xyz>0$，所以 $x>0$，$y>0$，$z>0$
$\begin{align}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\end{align}$
