\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y-12}-\dfrac{x}{y}=2\end{matrix}\right.\)
Đặt \(\dfrac{x}{y}=a;\dfrac{x}{y+12}=b;\dfrac{x}{y-12}=c\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\c-a=2\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{2}{a}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(\dfrac{1}{b}-\dfrac{1}{a}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
\(\Leftrightarrow\dfrac{a-b}{ab}+\dfrac{a-c}{ac}=0\Leftrightarrow\dfrac{1}{ab}-\dfrac{2}{ac}=0\)
\(\Leftrightarrow\dfrac{1}{a}\left(\dfrac{1}{b}-\dfrac{2}{c}\right)=0\Rightarrow\dfrac{1}{b}=\dfrac{2}{c}\Rightarrow c=2b\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\2b-a=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=4\\\dfrac{x}{y+12}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4y\\\dfrac{4y}{y+12}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=144\\y=36\end{matrix}\right.\)
Vậy . . .
