Đáp án: $A = 9$
Giải thích các bước giải:
$\begin{array}{l}
A = \dfrac{1}{{\sqrt 1 + \sqrt 2 }} + \dfrac{1}{{\sqrt 2 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \dfrac{1}{{\sqrt {99} + \sqrt {100} }}\\
= \dfrac{{\sqrt 2 - \sqrt 1 }}{{\left( {\sqrt 2 - \sqrt 1 } \right)\left( {\sqrt 2 + \sqrt 1 } \right)}} + \dfrac{{\sqrt 3 - \sqrt 2 }}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}\\
+ \dfrac{{\sqrt 4 - \sqrt 3 }}{{\left( {\sqrt 4 - \sqrt 3 } \right)\left( {\sqrt 4 + \sqrt 3 } \right)}} + ... + \dfrac{{\sqrt {100} - \sqrt {99} }}{{\left( {\sqrt {100} - \sqrt {99} } \right)\left( {\sqrt {100} + \sqrt {99} } \right)}}\\
= \dfrac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \dfrac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}} + \dfrac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}} + ... + \dfrac{{\sqrt {100} - \sqrt {99} }}{{100 - 99}}\\
= \sqrt 2 - 1 + \sqrt 3 - \sqrt 2 + \sqrt 4 - \sqrt 3 + ... + \sqrt {100} - \sqrt {99} \\
= \sqrt {100} - 1\\
= 10 - 1\\
= 9
\end{array}$
