Question

giải hệ phương trình \(\left\{{}\begin{matrix}2x^2-xy+y^2=16\\x^2+xy=12\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2-3xy+3y^2=48\\4x^2+4xy=48\end{matrix}\right.\)

\(\Rightarrow2x^2-7xy+3y^2=0\)

\(\Leftrightarrow\left(2x-y\right)\left(x-3y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\x=3y\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}y=2x\\x^2+x.2x=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\x^2=4\end{matrix}\right.\) \(\Rightarrow...\)

TH2: \(\left\{{}\begin{matrix}x=3y\\\left(3y\right)^2+3y.y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y^2=1\end{matrix}\right.\) \(\Rightarrow...\)