\(A=\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2\left(x+1\right)}}-\frac{3}{2}=\sqrt{6}-\frac{3}{2}\)
\(A_{min}=\sqrt{6}-\frac{3}{2}\) khi \(\frac{3\left(x+1\right)}{2}=\frac{1}{x+1}\Leftrightarrow x=\sqrt{\frac{2}{3}}-1\)