\(\dfrac{3x^2 + 6x+10}{x^2 + 2x+3}\) \((1) \)
= \(\dfrac{3(x^2+2x+3)+1}{x^2+2x+3}\)
\(= 3+ \dfrac{1}{(x+1)^2 +2}\)
Ta có: \((x+1)^2 \) \(\ge\) \(0\)
\(<=> (x+1)^2 +2\)\(\ge\) \(2\)
\(<=> \dfrac{1}{(x+1)^2 +2}\) \(\le\) \(\dfrac{1}{2}\)
\(<=> 3 + \dfrac{1}{(x+1)^2 +2}\) \(\le\) \(\dfrac{7}{2}\)
Vậy (1) max = \(\dfrac{7}{2}\) \(<=> x = -1 \)