Question

Giải hệ phương trình

\(3x-y\left(1+5x\right)=0\)

\(x\left(1-5y\right)+y=0\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}3x-y-5xy=0\\x-5xy+y=0\left(1\right)\end{matrix}\right.\)

\(\Rightarrow3x-y-5xy=x-5xy+y\)

\(\Leftrightarrow2x=2y\)

\(\Leftrightarrow x=y\)

Thay vào (1):

\(2x-5x^2=0\)

\(\Leftrightarrow x\left(5x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=y=0\\x=y=\frac{2}{5}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;0\right);\left(\frac{2}{5};\frac{2}{5}\right)\)