ta có : x:\(\dfrac{x}{y}\)=\(\dfrac{1}{3}\)
->x.\(\dfrac{y}{x}\)=\(\dfrac{1}{3}\)
->y=\(\dfrac{1}{3}\)
->x-\(\dfrac{3}{\dfrac{1}{3}}\)=\(\dfrac{1}{2}\)
->x = \(\dfrac{19}{2}\)
Vậy......
\(\dfrac{x}{y}=\dfrac{1}{3}\)
\(\Rightarrow3x=y\)
Xảy ra khi
\(x=\dfrac{1}{3}y\)
\(x-\dfrac{3}{y}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{2x}{2}-\dfrac{2}{y}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{2x}{2}-\dfrac{1}{2}=\dfrac{2}{y}\)
\(\Rightarrow\dfrac{2x-1}{2}=\dfrac{2}{y}\)
\(\Rightarrow y\left(2x-1\right)=4\)
\(\Rightarrow y;2x-1\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
mà 1-2y lẻ :
\(\left\{{}\begin{matrix}2x-1=1\Rightarrow2x=2\Rightarrow x=1\\y=4\\2x-1=-1\Rightarrow2x=0\Rightarrow x=0\\y=-4\end{matrix}\right.\)
