Question

tìm x,y, z nguyên dương sao cho: \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}=3\)

Answer 1

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Ta có: \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}=3\)

\(\Leftrightarrow\dfrac{xyz}{z^2}+\dfrac{xyz}{y^2}+\dfrac{xyz}{x^2}=3\)

\(\Leftrightarrow xyz\left(\dfrac{1}{z^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2}\right)=3=1\cdot3=3\cdot1\) (Vì x,y,z dương)

TH1: \(\left\{{}\begin{matrix}xyz=1\\\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3\end{matrix}\right.\) \(\Rightarrow x=y=z=1\) (TM)

TH2: \(\left\{{}\begin{matrix}xyz=3\\\dfrac{1}{x^2}+\dfrac{1}{y^1}+\dfrac{1}{z^2}=1\end{matrix}\right.\) \(\Rightarrow x=y=z=\varnothing\)

Vậy x=y=z=1