\(A=\dfrac{6x+8}{x^2+1}\)

\(=\dfrac{\left(-9+6x-1\right)\left(9x^2+9\right)}{x^2+1}\)

\(=-\dfrac{\left(3x-1\right)^2}{x+1}+9\)

Vì \(-\dfrac{\left(3x-1\right)^2}{x^2+1}\le0\) nên \(-\dfrac{\left(3x-1\right)^2}{x^2+1}+9\le9\)

Dấu '' = '' xảy ra khi \(x=\dfrac{1}{3}\)

Vậy \(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

Ta có : 

\(A=\dfrac{6x+8}{x^2+1}\)

\(=\dfrac{\left(x^2+6x+9\right)-\left(x^2+1\right)}{x^2+1}\)

\(=\dfrac{\left(x+3\right)^2}{x^2+1}-1\)

Vì \(\left(x+3\right)^2\ge0\) nên \(\dfrac{\left(x+3\right)^2}{x^2+1}\)

nên \(\dfrac{\left(x+3\right)^2}{x^2+1}-1\ge-1\) hay \(A>-1\)

Dấu ' = ' xảy ra khi \(x=-3\)

Vậy \(A_{min}=-1\) khi \(x=-3\)

Ta có :