\(\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\\y\left(x-y\right)=-\dfrac{3}{10}\end{matrix}\right.\)
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 \(x=\mp\sqrt{\dfrac{3}{20}};y=\pm\sqrt{\dfrac{3}{20}}\)
