Ta có :
\(3x^2+4x+8=3\left(x^2+\dfrac{4}{3}x+\dfrac{8}{3}\right)=3\left(x^2+2.\dfrac{2}{3}x+\dfrac{4}{9}\right)+8-\dfrac{4}{3}=\left(x+\dfrac{2}{3}\right)^2+\dfrac{20}{3}\ge\dfrac{20}{3}\)
\(\Rightarrow3x^2+4x+8\ge\dfrac{20}{3}\left(tại.x=-\dfrac{2}{3}\right)\)
Ta lại có :
\(x^2+3\ge3\left(tại.x=0\right)\)
- Với \(x=-\dfrac{2}{3}\)
\(A=\dfrac{3x^2+4x+8}{x^2+3}\le\dfrac{\dfrac{20}{3}}{\dfrac{4}{9}+3}=\dfrac{\dfrac{20}{3}}{\dfrac{31}{9}}=\dfrac{60}{31}\left(1\right)\)
- Với \(x=0\)
\(A=\dfrac{3x^2+4x+8}{x^2+3}\le\dfrac{8}{3}\left(2\right)\)
\(\left(1\right)\&\left(2\right)\Rightarrow A=\dfrac{3x^2+4x+8}{x^2+3}\le\dfrac{8}{3}\)
\(\Rightarrow GTLN\left(A\right)=\dfrac{8}{3}\left(tại.x=0\right)\)
