a) \(3=\sqrt{9}\) > \(\sqrt{7}\)

=> \(3\) > \(\sqrt{7}\)

b) +) \(5\sqrt{2}=\sqrt{50}\)

+)\(2\sqrt{5}=\sqrt{20}\)

mà \(\sqrt{50}>\sqrt{20}\)

=> \(5\sqrt{2}>2\sqrt{5}\)

c) +) \(7=3+4\) \(=\sqrt{9}+\sqrt{16}\)

vì \(\sqrt{9}+\sqrt{16}>\sqrt{7}+\sqrt{15}\)

=> \(\sqrt{7}+\sqrt{15}< 7\)

d) +) \(6-\sqrt{15}=\sqrt{36}-\sqrt{15}\)

vì \(\sqrt{36}-\sqrt{15}< \sqrt{37}-\sqrt{14}\)

=> \(\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)

e) +) 6 + \(2\sqrt{2}\) = \(6+\sqrt{8}\)

+) 6 + 3 = \(6+\sqrt{9}\)

vì 6 + \(\sqrt{8}\) < 6 + \(\sqrt{9}\)

=> 6 + \(2\sqrt{2}\) <\(6+3\)

A=\(\dfrac{1}{\sqrt{x-3}}\)

Để A xác định thì x - 3 > 0

\(\Leftrightarrow\) x > 3