Giải thích các bước giải:
$y=\dfrac{x^2+4x+5}{x^2+3x+3}$
$+) y-\dfrac{2}{3}=\dfrac{x^2+4x+5}{x^2+3x+3}-\dfrac{2}{3}=\dfrac{(x+3)^2}{3(x^2+3x+3)}\ge 0$
$\to y\ge \dfrac 23\to Min y=\dfrac 23\to x=-3$
$+) y-2=\dfrac{x^2+4x+5}{x^2+3x+3}-2=\dfrac{-(x+1)^2}{x^2+3x+3}\le 0$
$\to y\le 2\to Max y=2\to x=-1$
