Question

Tìm GTLN, GTNN của S=\(\dfrac{x+2}{x^2+x+1}\)

Answer 1

\(S=\dfrac{x+2}{x^2+x+1}\Rightarrow Sx^2+Sx+S=x+2\)

\(\Rightarrow Sx^2+\left(S-1\right)x+S-2=0\)

\(\Delta=\left(S-1\right)^2-4S\left(S-2\right)\ge0\)

\(\Leftrightarrow-3S^2+6S+1\ge0\)

\(\Rightarrow\dfrac{3-2\sqrt{3}}{3}\le S\le\dfrac{3+2\sqrt{3}}{3}\)