Tính giá trị của \(A =\dfrac{1}{{2\sqrt 1  + 1\sqrt 2 }} + \dfrac{1}{{3\sqrt 2  + 2\sqrt 3 }} + ... + \dfrac{1}{{2018\sqrt {2017}  + 2017\sqrt {2018} }}\)

Ta có: \(k\sqrt {k - 1}  + \left( {k - 1} \right)\sqrt k \, = \sqrt {k\left( {k - 1} \right)} \left( {\sqrt k  + \sqrt {k - 1} } \right)\) với \(k \ge 1\).

\(\begin{array}{l} \Rightarrow \dfrac{1}{{k\sqrt {k - 1}  + \left( {k - 1} \right)\sqrt k }} \\= \dfrac{1}{{\sqrt {k\left( {k - 1} \right)} \left( {\sqrt k  + \sqrt {k - 1} } \right)}} \\= \dfrac{{\left( {\sqrt k  - \sqrt {k - 1} } \right)}}{{\sqrt {k\left( {k - 1} \right)} \left( {\sqrt k  + \sqrt {k - 1} } \right)\left( {\sqrt k  - \sqrt {k - 1} } \right)}}\\ = \dfrac{{\sqrt k  - \sqrt {k - 1} }}{{\sqrt {k\left( {k - 1} \right)} }} \\= \dfrac{{\sqrt k  - \sqrt {k - 1} }}{{\sqrt k .\sqrt {k - 1} }} \\= \dfrac{1}{{\sqrt {k - 1} }} - \dfrac{1}{{\sqrt k }}\end{array}\)  

Thay lại vào A ta được:

\(A = \dfrac{1}{{2\sqrt 1  + 1\sqrt 2 }} + \dfrac{1}{{3\sqrt 2  + 2\sqrt 3 }}\)\( + ... + \dfrac{1}{{2018\sqrt {2017}  + 2017\sqrt {2018} }}\)\(= \,\left( {\dfrac{1}{{\sqrt 1 }} - \dfrac{1}{{\sqrt 2 }}} \right) + \left( {\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 3 }}} \right) \)\(+ ..... + \left( {\dfrac{1}{{\sqrt {2017} }} - \dfrac{1}{{\sqrt {2018} }}} \right)\)\(= 1 - \dfrac{1}{{\sqrt {2018} }}\)