ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(\frac{x^2+1}{y}\right)+2\left(x+y\right)=8\\\left(x+y\right)^2-2\left(\frac{x^2+1}{y}\right)=7\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2+2\left(x+y\right)=15\)
\(\Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)-15=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=3\Rightarrow\frac{x^2+1}{y}=1\\x+y=-5\Rightarrow\frac{x^2+1}{y}=9\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=3\\\frac{x^2+1}{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\x^2+1=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+x^2+1=3\\y=x^2+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-2=0\\y=x^2+1\end{matrix}\right.\) (casio)
TH2: \(\left\{{}\begin{matrix}x+y=-5\\\frac{x^2+1}{y}=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-5\\\frac{x^2+1}{9}=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{x^2+1}{9}=-5\\y=\frac{x^2+1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+9x+46=0\\y=\frac{x^2+1}{9}\end{matrix}\right.\) (vô nghiệm)
