Question

Giải hệ phương trình

\(\left\{{}\begin{matrix}2x^2+3y^2+xy-2y-4=0\\3x^2+4x+5y^2-12=0\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}2x^2+3y^2+xy-2y-4=0\left(1\right)\\3x^2+4x+5y^2-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+6y^2+2xy-4y-8=0\left(1'\right)\\3x^2+4x+5y^2-12=0\left(2'\right)\end{matrix}\right.\)

\(\left(1'\right)+\left(2'\right)\Leftrightarrow\left(x+y-2\right)^2=0\)

\(\Leftrightarrow x=2-y\) thay vào (1)

\(2\left(2-y\right)^2+3y^2+\left(2-y\right)y-2y-4=0\)

\(\Leftrightarrow4\left(y-1\right)^2=0\Leftrightarrow y=1\)

Khi dok \(\Leftrightarrow x=2-1=1\)

Vay x=y=1 la nghiem cua hpt