Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(\dfrac{2014}{\sqrt{2015}}+\dfrac{2015}{\sqrt{2014}}>\sqrt{2014}+\sqrt{2015}\)
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....
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 \(\frac{2014}{\sqrt{2015}}\)+ \(\frac{2015}{\sqrt{2014}}\) > \(\sqrt{2014} + \sqrt{2015}\)
\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}\)
\(=\sqrt{1+a^2+\left(\dfrac{a}{a+1}\right)^2+2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}}\)
(vì \(2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}=\dfrac{2a^2+2a-2a-2a^2}{a+1}=0\))
\(=\sqrt{\left(1+a-\dfrac{a}{a+1}\right)^2}\)
\(=\left|1+a-\dfrac{a}{a+1}\right|\)
- - -
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2014^2+\dfrac{2014^2}{2015^2}}+\dfrac{2014}{2015}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=\left|1+2014-\dfrac{2014}{2015}\right|+\dfrac{2014}{2015}\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2015\)
Tới đây bn làm bảng xét dấu nhé ~^^~
