Question

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

Answer 1

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

Nhân vế với vế:

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

\(\Leftrightarrow x^3+5x^2y+3xy^2-y^3=0\)

\(\Leftrightarrow\left(x+y\right)\left(x^2+4xy-y^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-x\\y=\left(2-\sqrt{5}\right)x\\y=\left(2+\sqrt{5}\right)x\end{matrix}\right.\)

Thay vào pt dưới ...