Ta có :

\(\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\left(1\right)\\y\left(x-y\right)=\dfrac{-3}{50}\left(2\right)\end{matrix}\right.\)

Lấy vế \(\left(1\right)\) chia theo vế cho \(\left(2\right)\) ta được :

\(\dfrac{x\left(x-y\right)}{y\left(x-y\right)}=\dfrac{x}{y}=\dfrac{3}{10}:\dfrac{-3}{50}=-5\)

\(\Rightarrow\dfrac{x}{y}=-5\Rightarrow x=-5y\left(3\right)\)

Thay \(\left(3\right)\) vào \(\left(2\right)\) Ta có :

\(y\left(-5y-y\right)=\dfrac{-3}{50}\)

\(\Rightarrow-6y^2=\dfrac{-3}{50}\)

Bn tự tính típ nhs!

\(\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\\y\left(x-y\right)=-\dfrac{3}{10}\end{matrix}\right.\)

Cộng theo vế 2 pt trên ta có:

\(x\left(x-y\right)+y\left(x-y\right)=\dfrac{-3}{10}+\dfrac{3}{10}=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)=0\)\(\Rightarrow\left[{}\begin{matrix}x=-y\\x=y\end{matrix}\right.\)

*)Xét \(x=-y\) thì \(x\left(x-y\right)=\dfrac{3}{10}\Leftrightarrow-y\left(-y-y\right)=\dfrac{3}{10}\)

\(\Leftrightarrow2y^2=\dfrac{3}{10}\Leftrightarrow y^2=\dfrac{3}{20}\Rightarrow y=\pm\sqrt{\dfrac{3}{20}}\)\(\Rightarrow x=\mp\sqrt{\dfrac{3}{20}}\)

*)Xét \(x=y\) thì \(x\left(x-y\right)=\dfrac{3}{10}\Leftrightarrow y\left(y-y\right)=\dfrac{3}{10}\)

\(\Leftrightarrow0=\dfrac{3}{10}\) (loại)

Vậy \(x=\mp\sqrt{\dfrac{3}{20}};y=\pm\sqrt{\dfrac{3}{20}}\)