Lời giải:
Có \(y=\frac{x^2+2x+2}{x^2+2}\Leftrightarrow x^2y+2y=x^2+2x+2\)
\(\Leftrightarrow x^2(y-1)-2x+(2y-2)=0\)\((\star)\)
Nếu \(y-1=0\rightarrow y=1(1)\Leftrightarrow \frac{x^2+2x+2}{x^2+2}=1+\frac{2x}{x^2+2}=1\Leftrightarrow x=0\)
Nếu \(yeq 1\). Điều kiện để PT \((\star)\) có nghiệm là:
\(\Delta'=1-(y-1)(2y-2)\geq 0\)
\(\Leftrightarrow (y-1)^2\leq \frac{1}{2}\Rightarrow \frac{-1}{\sqrt{2}}\leq y-1\leq \frac{1}{\sqrt{2}}\)
\(\Leftrightarrow 1-\frac{1}{\sqrt{2}}\leq y\leq 1+\frac{1}{\sqrt{2}}\)\((2)\)
Từ \((1),(2)\Rightarrow \)\(\left\{\begin{matrix} y_{\min}=1-\frac{1}{\sqrt{2}}\Leftrightarrow x=-\sqrt{2}\\ y_{\max}=1+\frac{1}{\sqrt{2}}\Leftrightarrow x=\sqrt{2}\end{matrix}\right.\)
