\(\hept{\begin{cases}3x-y\left(5x+1\right)=0\\x\left(1-5y\right)+y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x-5xy-y=0\\x-5xy+y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4x-10xy=0\\x-5xy+y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4x\left(1-\frac{5}{2}y\right)=0\\x-5xy+y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0;\frac{5}{2}y=1\\x-5xy+y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=\frac{2}{5}\\-5.\frac{2}{5}x+\frac{2}{5}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=\frac{2}{5}\\x=\frac{1}{5}\end{cases}}\)
vậy hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(\frac{1}{5};\frac{2}{5}\right)\)
