Question

Giải hệ phương trình:

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

Answer 1

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

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

Trừ vế cho vế:

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

\(\Delta=\left(x^2-x+3\right)^2-4\left(-x^3+2x^2-x+2\right)=\left(x^2+x-1\right)^2\)

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

Thế vào pt dưới:

\(\left[{}\begin{matrix}x+x^2+1=2\\x-x+2=\dfrac{x^2+1-x+2}{x^2+1}\end{matrix}\right.\)

\(\Leftrightarrow...\)