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c: Ta có: \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)
\(=4+\sqrt{10}-4+\sqrt{10}\)
\(=2\sqrt{10}\)
d: Ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)
\(=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1\)
\(=2\sqrt{2}\)
a) \(=\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2=12-18=-6\)
b) \(=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}-\sqrt{2015}=-\sqrt{2013}-\sqrt{2015}\)
c) \(=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
d) \(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)
Áp dụng bất đẳng thức Cauchy, ta có:
\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)
Cộng vế theo vế, ta được:
\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)
Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)
Vậy....
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
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 \(\frac{2014}{\sqrt{2015}}\)+ \(\frac{2015}{\sqrt{2014}}\) > \(\sqrt{2014} + \sqrt{2015}\)
\(\frac{1}{\sqrt{2013}-\sqrt{2014}}-\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
\(=\frac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2014}-\sqrt{2015}\right)}-\frac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}\)
\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)
\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)
\(=\sqrt{2015}-\sqrt{2013}\)
\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)
\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)
\(=\sqrt{2015}-\sqrt{2013}\)
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