Tìm cặp giá trị \((m;n)\)  để hai hệ phương trình sau tương đương \(\left\{ \begin{array}{l}3x + 3y = 3\\x + \dfrac{1}{3}y = \dfrac{1}{3}\end{array} \right.(I)\) và

$\left\{ \begin{array}{l}{\rm{x}} - ny = 1\\3mx + my = 1\end{array} \right.(II)$

\(\left\{ \begin{array}{l}y = 3x\\\sqrt x  + \sqrt y  = 1\end{array} \right.\)

\(\left\{ \begin{array}{l}x - 5y = 3\\y = 3x - 1\end{array} \right.\)

\(\left\{ \begin{array}{l}3 + 2y = 5x\\{x^3} = {y^2}\end{array} \right.\)

\(\left\{ \begin{array}{l}x + 2y + z = 10\\ - 2x + 3y + z = 2019\end{array} \right.\)

\(\left\{ \begin{array}{l}x - 2y =  - 1\\3x - y = 4\end{array} \right.\)

\(\left\{ \begin{array}{l}{x^2} - 3y =  - 4\\y - 3\sqrt 5 x = 4\end{array} \right.\)

\(\left\{ \begin{array}{l}5x + 6y = 19\\ - x - y =  - 3\end{array} \right.\)

\(\left\{ \begin{array}{l}x + \sqrt y  =  - 3\\4x + y = \sqrt 5 \end{array} \right.\)

\(\left( { - \dfrac{1}{6}; - \dfrac{5}{3}} \right)\)     

\(\left( {\dfrac{1}{3};\dfrac{5}{6}} \right)\)

\(\left( { - \dfrac{1}{3};\dfrac{5}{6}} \right)\)    

\(\left( { - \dfrac{1}{3}; - \dfrac{5}{6}} \right)\)

\(\left( { - 1;1} \right)\)

\(\left( { - 1; - 1} \right)\)

\(\left( {1; - 1} \right)\)

\(\left( {1;1} \right)\)

\(a = 1\) 

\(a =  - 1\)       

\(a =  - 2\)          

\(a = 2\)