Đặt \(u=\sqrt{x-2}\)
+) u > 0
\(+)x=u^2+2\Rightarrow A=\frac{\left(u^2+2\right)+3u}{\left(u^2+2\right)+4u+1}=\frac{u^2+3u+2}{u^2+4u+3}\)
\(=\frac{\left(u+1\right)\left(u+2\right)}{\left(u+1\right)\left(u+3\right)}=\frac{u+2}{u+3}=1-\frac{1}{u+3}\)
+) Vì \(u\ge0\)nên \(u+3\ge3\)
\(\Rightarrow\frac{1}{u+3}\le\frac{1}{3}\)hay \(-\frac{1}{u+3}\ge-\frac{1}{3}\)
\(\Rightarrow A\ge1-\frac{1}{3}=\frac{2}{3}\)
+) Khi x = 2 thì \(A=\frac{2}{3}\)
Vậy min \(A=\frac{2}{3}\)