Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}\le\sqrt{\frac{a+b}{2}}\) :

Xét : \(N-M=2\sqrt{2014}-\left(\sqrt{2015}+\sqrt{2013}\right)\)

Theo bđt trên thì \(\frac{\sqrt{2013}+\sqrt{2015}}{2}\le\sqrt{\frac{2013+2015}{2}}\Leftrightarrow\sqrt{2013}+\sqrt{2015}\le2\sqrt{2014}\)

\(\Rightarrow N-M>0\Rightarrow N>M\)

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Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{2015-2014}{\sqrt{2015}+\sqrt{2014}}>\dfrac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\sqrt{2016}-\sqrt{2015}\)

Ta có:  √2015−√2014=2015−2014√2015+√2014>2016−2015√2016+√2015=√2016−√2015

a) Ta có: \(2\sqrt{3}=\sqrt{4\cdot3}=\sqrt{12}\)

\(3\sqrt{2}=\sqrt{9\cdot2}=\sqrt{18}\)

mà \(\sqrt{12}< \sqrt{18}\)(vì 12<18)

nên \(2\sqrt{3}< 3\sqrt{2}\)

b) Ta có: \(\left(2\sqrt{3}+1\right)^2=8+4\sqrt{3}+1=9+4\sqrt{3}\)

\(4^2=16=9+7\)

mà \(4\sqrt{3}< 7\left(\sqrt{48}< \sqrt{49}\right)\)

nên \(\left(2\sqrt{3}+1\right)^2< 4^2\)

hay \(2\sqrt{3}+1< 4\)

c) Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)

\(\sqrt{2014}-\sqrt{2013}=\dfrac{1}{\sqrt{2014}+\sqrt{2013}}\)

Ta có: \(\sqrt{2015}+\sqrt{2014}>\sqrt{2013}+\sqrt{2014}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{2015}+\sqrt{2014}}< \dfrac{1}{\sqrt{2013}+\sqrt{2014}}\)

hay \(\sqrt{2015}-\sqrt{2014}< \sqrt{2014}-\sqrt{2013}\)

\(a\))Ta có:\(2\sqrt{3}=\sqrt{12}\)

             \(3\sqrt{2}=\sqrt{18}\)

Vì \(\sqrt{12}< \sqrt{18}\)

⇒\(2\sqrt{3}< 3\sqrt{2}\)

\(b\))Ta có:\(2\sqrt{3}+1=\sqrt{12}+1\)

             \(4=3+1=\sqrt{9}+1\)

Vì \(\sqrt{12}+1>\sqrt{9}+1\)

⇒\(2\sqrt{3}+1>4\)