a) Chứng minh rằng \(\frac{1}{{\sqrt {n + 1} + \sqrt n }} = \sqrt {n + 1} - \sqrt n \) với mọi số tự nhiên n.
b) Tính \(\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}.\)
Dựa vào: \(\frac{{\sqrt a }}{{\sqrt b }} = \frac{{\sqrt a .\sqrt b }}{{{{\left( {\sqrt b } \right)}^2}}} = \frac{{\sqrt {ab} }}{b}(a \ge 0,b > 0)\)
\(\sqrt {\frac{a}{b}} = \sqrt {\frac{{ab}}{{{b^2}}}} = \frac{{\sqrt {ab} }}{b}(a \ge 0,b > 0)\)
Advertisements (Quảng cáo)
a) Xét vế trái
\(VT = \frac{1}{{\sqrt {n + 1} + \sqrt n }} \\= \frac{{\sqrt {n + 1} - \sqrt n }}{{\left( {\sqrt {n + 1} + \sqrt n } \right)\left( {\sqrt {n + 1} - \sqrt n } \right)}}\\ = \frac{{\sqrt {n + 1} - \sqrt n }}{{n + 1 - n}} \\= \sqrt {n + 1} - \sqrt n = VP.\)
b) Ta có:
\(\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }} \\= \frac{\sqrt 1 - \sqrt 2}{{(\sqrt 1 + \sqrt 2 )(\sqrt 1 - \sqrt 2)}} + \frac{\sqrt 2 - \sqrt 3}{{(\sqrt 2 + \sqrt 3)(\sqrt 2 - \sqrt 3)}} + ... + \frac{\sqrt {99} - \sqrt {100} }{{(\sqrt {99} + \sqrt {100})(\sqrt {99} - \sqrt {100}) }} \\= \frac{\sqrt 1 - \sqrt 2}{{1 - 2}} + \frac{\sqrt 2 - \sqrt 3}{{2 - 3}} + ... + \frac{\sqrt {99} - \sqrt {100} }{{99 - 100 }} \\= - (\sqrt 1 - \sqrt 2) - (\sqrt 2 - \sqrt 3) - ... - (\sqrt {99} - \sqrt {100}) \\= \sqrt 2 - 1 + \sqrt 3 - \sqrt 2 + ... + \sqrt {100} - \sqrt {99} \\ = - 1 + \sqrt {100} = - 1 + 10 = 9.\)