ĐKXĐ: ...
Đặt \(\left(x+\frac{2}{x}\right)^2=a\ge8\Rightarrow x^2+\frac{4}{x^2}=a-4\)
\(\Rightarrow A=\sqrt{\left(a-4\right)^2-8a+50}\)
\(A=\sqrt{a^2-16a+66}\)
\(A=\sqrt{\left(a-8\right)^2+2}\ge\sqrt{2}\)
\(A_{min}=\sqrt{2}\) (khi \(a=8\) hay \(x=\pm\sqrt{2}\))