Đáp án:
\(1)A = - 1 + \sqrt {2021} \)
Giải thích các bước giải:
\(\begin{array}{l}
1)A = \dfrac{{\sqrt 1 - \sqrt 2 }}{{\left( {\sqrt 1 - \sqrt 2 } \right)\left( {\sqrt 1 + \sqrt 2 } \right)}} + \dfrac{{\sqrt 2 - \sqrt 3 }}{{\left( {\sqrt 2 - \sqrt 3 } \right)\left( {\sqrt 2 + \sqrt 3 } \right)}} + ... + \dfrac{{\sqrt {2020} - \sqrt {2021} }}{{\left( {\sqrt {2020} - \sqrt {2021} } \right)\left( {\sqrt {2020} + \sqrt {2021} } \right)}}\\
= \dfrac{{\sqrt 1 - \sqrt 2 }}{{1 - 2}} + \dfrac{{\sqrt 2 - \sqrt 3 }}{{2 - 3}} + ... + \dfrac{{\sqrt {2020} - \sqrt {2021} }}{{2020 - 2021}}\\
= \dfrac{{\sqrt 1 - \sqrt 2 }}{{ - 1}} + \dfrac{{\sqrt 2 - \sqrt 3 }}{{ - 1}} + ... + \dfrac{{\sqrt {2020} - \sqrt {2021} }}{{ - 1}}\\
= - \sqrt 1 + \sqrt 2 - \sqrt 2 + \sqrt 3 - ... - \sqrt {2020} + \sqrt {2021} \\
= - 1 + \sqrt {2021} \\
2)B = \dfrac{{\sqrt 1 + \sqrt 2 }}{{\left( {\sqrt 1 - \sqrt 2 } \right)\left( {\sqrt 1 + \sqrt 2 } \right)}} - \dfrac{{\sqrt 2 + \sqrt 3 }}{{\left( {\sqrt 2 - \sqrt 3 } \right)\left( {\sqrt 2 + \sqrt 3 } \right)}} + ... + \dfrac{{\sqrt {99} + \sqrt {100} }}{{\left( {\sqrt {99} + \sqrt {100} } \right)\left( {\sqrt {99} - \sqrt {100} } \right)}}\\
= \dfrac{{\sqrt 1 + \sqrt 2 }}{{1 - 2}} - \dfrac{{\sqrt 2 + \sqrt 3 }}{{2 - 3}} + ... + \dfrac{{\sqrt {99} + \sqrt {100} }}{{99 - 100}}\\
= - 1 - \sqrt 2 + \sqrt 2 + \sqrt 3 - ... - \sqrt {99} - \sqrt {100} \\
= - 1 - \sqrt {100}
\end{array}\)
