Chứng minh A < 18.

Ta có:

\[\frac{1}{{2\sqrt 2 }} < \frac{1}{{\sqrt 1 + \sqrt 2 }} = \frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} = \sqrt 2 - \sqrt 1 .\]

\[\frac{1}{{2\sqrt 3 }} < \frac{1}{{\sqrt 2 + \sqrt 3 }} = \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}} = \sqrt 3 - \sqrt 2 .\]

\[\frac{1}{{2\sqrt 4 }} < \frac{1}{{\sqrt 3 + \sqrt 4 }} = \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}} = \sqrt 4 - \sqrt 3 .\]

...

\[\frac{1}{{2\sqrt {100} }} = \frac{1}{{\sqrt {99} + \sqrt {100} }} = \frac{{\sqrt {100} - \sqrt {99} }}{{100 - 99}} = \sqrt {100} - \sqrt {99} .\]

Suy ra \[\frac{A}{2} < \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 + \sqrt 4 - \sqrt 3 + ... + \sqrt {100} - \sqrt {99} = \sqrt {100} - \sqrt 1 = 10 - 1 = 9.\]

Do đó A < 18.

Vậy A < 18.