Question

[thỏ]

Tìm các số tự nhiên x,y,z sao cho 0<x\(\le y\le z\) và xy+yz+zx= xyz

Answer 1

[Hoang Hung Quan]

Do \(x,y,z>0\Rightarrow xyz\ne0\)

\(\Rightarrow\dfrac{xy}{xyz}+\dfrac{yz}{xyz}+\dfrac{zx}{xyz}=1\)

\(\Rightarrow\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}=1\Rightarrow\dfrac{1}{x}< 1\Rightarrow x>1\)

Vì \(x\le y\le z\Rightarrow\dfrac{1}{x}\ge\dfrac{1}{y}\ge\dfrac{1}{z}\)

\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}=\dfrac{3}{x}\)

\(\Rightarrow1\le\dfrac{3}{x}\Rightarrow x\le3\) Mà \(x>1\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Nếu \(x=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2}\Rightarrow\dfrac{1}{y}< \dfrac{1}{2}\Rightarrow y>2\\\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{2}{y}\Rightarrow\dfrac{2}{y}\ge\dfrac{1}{2}\Rightarrow y\le4\end{matrix}\right.\)

Mà \(y>2\Rightarrow\left[{}\begin{matrix}y=3\\y=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=3\Rightarrow z=6\\y=4\Rightarrow z=4\end{matrix}\right.\)

Nếu \(x=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{2}{3}\Rightarrow\dfrac{1}{y}< \dfrac{2}{3}\Rightarrow y>\dfrac{3}{2}\\\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{2}{y}\Rightarrow\dfrac{2}{y}\ge\dfrac{2}{3}\Rightarrow y\le3\end{matrix}\right.\)

Do \(x\le y\Rightarrow\left\{{}\begin{matrix}y=3\\z=3\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(3;3;3\right);\left(2;3;6\right);\left(2;4;4\right)\)

Answer 2

[thỏ]

giúp nha, đúng mình tick cho