\(\Leftrightarrow\left\{{}\begin{matrix}3x-y-5xy=0\\x-5xy+y=0\left(1\right)\end{matrix}\right.\)
\(\Rightarrow3x-y-5xy=x-5xy+y\)
\(\Leftrightarrow2x=2y\)
\(\Leftrightarrow x=y\)
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 \(\left(x;y\right)=\left(0;0\right);\left(\frac{2}{5};\frac{2}{5}\right)\)
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