$A=\dfrac{2x+1}{x^2+2}$

*Min A:

Ta có: $A=\dfrac{2x+1}{x^2+2}$

$=\dfrac{4x+2}{2\left(x^2+2\right)}=\dfrac{\left(x^2+4x+4\right)-\left(x^2+2\right)}{2\left(x^2+2\right)}$

$=\dfrac{\left(x+2\right)^2}{2\left(x^2+1\right)}+\dfrac{1}{2}\ge\dfrac{1}{2},\forall x\in R$

Vậy $Min_A=\dfrac{1}{2}khi\left(x+2\right)^2=0$

$\Leftrightarrow x+2=0\Leftrightarrow x=-2$

*Max A:

Ta có: $A=\dfrac{2x+1}{x^2+2}$

$=\dfrac{x^2+2-x^2+2x-1}{x^2+2}$

$=\dfrac{(x^2+2)-(x^2-2x+1)}{x^2+2}$

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

$=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le0,\forall x\in R$

Vậy $Max_A=1khi\left(x-1\right)^2=0$

$\Leftrightarrow x-1=0\Leftrightarrow x=1$