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