$y=\dfrac{x^2+x+1}{x^2+1}$
$\to (x^2+1)y=x^2+x+1$
$\to x^2y+y-x^2-x-1=0$
$\to (y-1)x^2-x+y-1=0$ $(*)$
Hàm số $y=f(x)=\dfrac{x^2+x+1}{x^2+1}$ xác định với mọi $x\in\mathbb{R}$ do $x^2+1\ge 1>0\quad\forall x\in\mathbb{R}$
$\to (*)$ luôn có nghiệm với mọi $y$
$\Delta=1-4(y-1)^2\ge 0$
$\to (y-1)^2\le \dfrac{1}{4}$
$\to -\dfrac{1}{2}\le y-1\le \dfrac{1}{2}$
$\to \dfrac{1}{2}\le y\le \dfrac{3}{2}$
Vậy:
$\max y=\dfrac{3}{2}\Leftrightarrow x=1$
$\min y=\dfrac{1}{2}\Leftrightarrow x=-1$
