Đặt \(\sqrt{2x-1}=t\ge0\Rightarrow x=\frac{t^2+1}{2}\)
\(A=\frac{2\left(t^2+3t+2\right)}{t^2+4t+3}=\frac{2\left(t+1\right)\left(t+2\right)}{\left(t+1\right)\left(t+3\right)}=\frac{2\left(t+2\right)}{t+3}=2-\frac{2}{t+3}\ge2-\frac{2}{3}=\frac{4}{3}\)
\(A_{min}=\frac{4}{3}\) khi \(t=0\) hay \(x=\frac{1}{2}\)