\(\left\{{}\begin{matrix}x^2+\left(x+y\right)y+2=9y\\x+y-7=\dfrac{y}{x^2+2}\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+2\right)+\left(x+y-7\right)y=2y\\x+y-7=\dfrac{y}{x^2+2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+2=a\left(a\ge2\right)\\x+y-7=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+by=2y\\b=\dfrac{y}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2y-by\\ab=y\end{matrix}\right.\)
\(\Rightarrow\left(2y-by\right)b=y\).
Nhận thấy y = 0 không là nghiệm của hệ phương trình. Chia cả 2 vế cho y, ta được
\(\Rightarrow\left(2-b\right)b=1\)
\(\Leftrightarrow\left(b-1\right)^2=0\)
\(\Leftrightarrow b=1\)
\(\Rightarrow x+y-7=1\)
\(\Leftrightarrow y=8-x\). Thay vào (1)
\(\Rightarrow x+\left(8-x\right)-7=\dfrac{8-x}{x^2+2}\)
\(\Leftrightarrow1=\dfrac{8-x}{x^2+2}\)
\(\Rightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=6\\y=11\end{matrix}\right.\)
Vậy . . . =^~^=
