Question

Đã trả lời bởi chuyên gia

Answer 1

Lời giải

Ta có: \(\frac{1}{{\sqrt {k - 1} + \sqrt k }} > \frac{1}{{\sqrt k + \sqrt {k + 1} }}\)

\( \Rightarrow \frac{2}{{\sqrt {k - 1} + \sqrt k }} > \frac{1}{{\sqrt k + \sqrt {k + 1} }} + \frac{1}{{\sqrt {k - 1} + \sqrt k }}\)

\( = \sqrt k - \sqrt {k - 1} + \sqrt {k + 1} - \sqrt k = \sqrt {k + 1} - \sqrt {k - 1} \)

\( \Rightarrow 2\left( {\frac{1}{{1 + \sqrt 2 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {79} + \sqrt {80} }}} \right) > \sqrt 3 - \sqrt 1 + \sqrt 5 - \sqrt 3 + ... + \sqrt {81} - \sqrt {79} \)

\( \Rightarrow 2\left( {\frac{1}{{1 + \sqrt 2 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {79} + \sqrt {80} }}} \right) > 9 - 1 = 8\).

Vậy \[\frac{1}{{1 + \sqrt 2 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {79} + \sqrt {80} }} > 4\].

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Answer 2