\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2023}=\dfrac{1}{a+b+c}\\ \text{⇔}\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\\ \text{⇔}\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\\ \text{⇔}\left(a+b\right)\left(\dfrac{c\left(a+b+c\right)+ab}{abc\left(a+b+c\right)}\right)=0\\ \text{⇔}\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(a+b+c\right)}=0\\ \text{⇔}\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\)
$\Rightarrow (2023 - a)(2023 - b)(2023-c) = 0$
Vậy 1 trong ba số a,b,c bằng 2023
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