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Ta có : \(\frac{2014}{\sqrt{2015}}\)+ \(\frac{2015}{\sqrt{2014}}\) = \(\frac{2015-1}{\sqrt{2015}}\) + \(\frac{2014+1}{\sqrt{2014}}\)
= \(\sqrt{2015}\) + \(\sqrt{2014}\) + \(\frac{1}{\sqrt{2014}}\) - \(\frac{1}{\sqrt{2015}}\)
Vì \(\sqrt{2014}\) < \(\sqrt{2015}\) \(\Rightarrow \) \(\frac{1}{\sqrt{2014}}\)>\(\frac{1}{\sqrt{2015}}\) \(\Rightarrow \) \(\frac{1}{\sqrt{2014}}\)-\(\frac{1}{\sqrt{2015}}\) > 0
Nên \(\sqrt{2015}\) + \(\sqrt{2014}\) + \(\frac{1}{\sqrt{2014}}\) - \(\frac{1}{\sqrt{2015}}\) > \(\sqrt{2015}\) + \(\sqrt{2014}\)
Hay \(\frac{2014}{\sqrt{2015}}\)+ \(\frac{2015}{\sqrt{2014}}\) > \(\sqrt{2014} + \sqrt{2015}\)
Giải:
\(\dfrac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}\) \(=\dfrac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)
\(=\dfrac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\)
Áp dụng vào biểu thức ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}\) \(+...+\dfrac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2014}}-\dfrac{1}{\sqrt{2015}}\)
\(=1-\dfrac{1}{\sqrt{2015}}\)
Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014
\(\frac{1}{\sqrt{2013}-\sqrt{2014}}-\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
\(=\frac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2014}-\sqrt{2015}\right)}-\frac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}\)
\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)
\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)
\(=\sqrt{2015}-\sqrt{2013}\)
\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)
\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)
\(=\sqrt{2015}-\sqrt{2013}\)
Câu này mình từng làm qua (tuy khác đề) nhưng bạn chỉ cần áp dụng đúng công thức là OK nhé: https://olm.vn/hoi-dap/question/1294056.html
a) Ta có: \(\dfrac{2014}{\sqrt{2015}}+\dfrac{2015}{\sqrt{2014}}=\)
\(\dfrac{2015-1}{\sqrt{2015}}+\dfrac{2014+1}{\sqrt{2014}}=\sqrt{2015}-\dfrac{1}{\sqrt{2015}}+\sqrt{2014}+\dfrac{1}{\sqrt{2014}}\)
\(\left(\dfrac{1}{\sqrt{2014}}-\dfrac{1}{\sqrt{2015}}>0\right)\)\(>\sqrt{2014}+\sqrt{2015}\)
Vậy \(\dfrac{2014}{\sqrt{2015}}+\dfrac{2015}{\sqrt{2014}}>\sqrt{2014}+\sqrt{2015}\)
Ta có:
\(\dfrac{1}{\sqrt{a}-\sqrt[]{a+1}}=\dfrac{\sqrt{a}+\sqrt{a+1}}{a-a+1}=\sqrt{a}+\sqrt{a+1}\)
\(\Rightarrow\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\sqrt{2013}+\sqrt{2014}-\sqrt{2014}-\sqrt{2015}=\sqrt{2013}-\sqrt{2015}\)
hinh nhu sai roiii