Question

\(\left\{{}\begin{matrix}6x^2-3xy+x=1-y\\x^2+y^2=1\end{matrix}\right.\)

Answer 1

\(6x^2+x-1-3xy+y=0\)

\(\Leftrightarrow\left(3x-1\right)\left(2x+1\right)-y\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(2x+1-y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\y=2x+1\end{matrix}\right.\)

- Với \(x=\frac{1}{3}\) thay xuống dưới:

\(\frac{1}{9}+y^2=1\Rightarrow y^2=\frac{8}{9}\Rightarrow y=...\)

- Với \(y=2x+1\) thay xuống dưới:

\(x^2+\left(2x+1\right)^2=1\Leftrightarrow5x^2+4x=0\Leftrightarrow...\)