ĐKXĐ: a ≥ 2
Ta có:
A= \(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)
= \(\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)
= \(\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)
= \(\sqrt{a-2}+2+\left|\sqrt{a-2}-2\right|\)
Nếu a ≥ 6 thì A = \(\sqrt{a-2}+2+\sqrt{a-2}-2\)= \(2\sqrt{a-2}\)
Nếu 2 ≤ a < 6 thì A = \(\sqrt{a-2}+2+2-\sqrt{a-2}\) = 4