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

Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) với a > 0; b > 0; a \(\ne\) b ta có:

\(\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{2016+2014}{2}}\)

\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{4030}{2}}\)

\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{2015}\)

\(\Rightarrow\sqrt{2016}+\sqrt{2014}< 2.\sqrt{2015}\)

\(\Rightarrow\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

a) \(\sqrt{2017}-2\sqrt{2016}=\sqrt{2017}-\sqrt{8064}< 0< \sqrt{2016}\)

b) \(\sqrt{10}+\sqrt{17}+1>\sqrt{9}+\sqrt{16}+1=8=\sqrt{64}>\sqrt{61}\)

c) \(\left(\sqrt{2016}+\sqrt{2014}\right)^2=4030+\sqrt{2014.2016}\)

\(\left(2\sqrt{2015}^2\right)=4030+\sqrt{2015.2015}\)

C/m được: \(\sqrt{2014.2016}< \sqrt{2015.2015}\)

\(\Rightarrow\left(\sqrt{2016}+\sqrt{2014}\right)^2< \left(2\sqrt{2015}\right)^2\)

\(\Rightarrow\sqrt{2014}+\sqrt{2016}< 2\sqrt{2015}\)

d) \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=7=8-1=\sqrt{64}-1< \sqrt{65}-1\)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

\(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)

mà \(\sqrt{2016}+\sqrt{2015}>\sqrt{2014}+\sqrt{2015}\)

nên \(\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

\(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\)

=> \(\frac{1}{\sqrt{2016}+\sqrt{2015}}

A = \(\frac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\frac{1}{\sqrt{2016}+\sqrt{2015}}\); B = \(\frac{2015-2014}{\sqrt{2015}+\sqrt{2014}}=\frac{1}{\sqrt{2015}+\sqrt{2014}}\)

Mà \(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\) ( Vì \(\sqrt{2016}>\sqrt{2014}\))

Nên \(\frac{1}{\sqrt{2016}+\sqrt{2015}}

Ta đặt \(x=2015\), khi đó \(2014=x-1,2016=x+1.\)  Ta có như sau

\(2014^2\cdot2016=\left(x-1\right)^2\left(x+1\right)=\left(x^2-1\right)\left(x-1\right)\)\(