a) < 2 n N* (với a = 1, b= :sqrt{k} , A = n)
b) (1 + :frac{1}{n})^k < 1 + :frac{k}{n} + :frac{k^2}{n^2}
Edited by hoahoctro123, 11-09-2009 - 22:15.
Edited by hoahoctro123, 11-09-2009 - 22:15.
a)$\dfrac{1}{\sqrt{k}}=\dfrac{2}{2\sqrt{k}}<\dfrac{2}{\sqrt{k}+\sqrt{k-1}}=2(\sqrt{k}-\sqrt{k-1})$CMR
a) < 2 n N* (với a = 1, b= :sqrt{k} , A = n)
b) (1 + :frac{1}{n})^k < 1 + :frac{k}{n} + :frac{k^2}{n^2}
Edited by AnSatTruyHinh, 12-09-2009 - 18:29.
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