Question

giải hệ phương trình

\(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0\end{cases}}\)

Answer 1

\(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0\end{cases}}\)

\(\Leftrightarrow2x^2+xy-y^2-5x+y+2=x^2+y^2+x+y-4\)

\(\Leftrightarrow x^2+xy-y^2-5x+y+2=y^2+x+y-4\)

\(\Leftrightarrow x^2+xy-y^2-5x+y=y^2+x+y-4-2\)

\(\Leftrightarrow x^2+xy-y^2-5x+y=y^2+x+y-6\)

\(\Leftrightarrow x^2+xy-y^2+y=y^2+x+y-6+5x\)

\(\Leftrightarrow x^2+xy-y^2+y=y^2+6x+y-6\)

\(\Leftrightarrow x^2+xy-y^2=y^2+6x-6\)

\(\Leftrightarrow x^2+xy=y^2+6x-6+y^2\)

\(\Leftrightarrow x^2+xy=2y^2+6x-6\)

\(\Leftrightarrow x\left(x+y\right)=2\left(y^2+3x-3\right)\)