Ta có \(\left(\sqrt{2017}+\sqrt{2015}\right)^2=2017+2015+2\sqrt{2017.2015}=2.2016+\sqrt{\left(2016+1\right)\left(2016-1\right)}=2.2016+\sqrt{2016^2-1^2}\)
\(\sqrt{2017}-\sqrt{2016}=\dfrac{\left(\sqrt{2017}-\sqrt{2016}\right)\left(\sqrt{2017}+\sqrt{2016}\right)}{\sqrt{2017}+\sqrt{2016}}\\ =\dfrac{2017-2016}{\sqrt{2017}+\sqrt{2016}}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{\left(\sqrt{2016}-\sqrt{2015}\right)\left(\sqrt{2016}+\sqrt{2015}\right)}{\sqrt{2016}+\sqrt{2015}}\\ =\dfrac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
Mà:
\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\\ \Leftrightarrow\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\\ \Leftrightarrow\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
Tiếp phần trước
Ta có \(\left(2\sqrt{2016}\right)^2=4.2016=2.2016+2.2016=2.2016+2.\sqrt{2016}^2\)
Vì \(-1\) nhỏ hơn 0
\(\Rightarrow2016^2-1\) nhỏ hơn \(2016^2\)
\(\Rightarrow\sqrt{2016^2-1}\) nhỏ hơn \(\sqrt{2106^2}\)
\(\Rightarrow2.\sqrt{2016^2-1}\)nhỏ hơn\(\sqrt{2016^2}\)
\(\Rightarrow\left(\sqrt{2015}+\sqrt{2017}\right)^2\) nhỏ hơn \(\left(2\sqrt{2016}\right)^2\)
\(\Rightarrow\sqrt{2017}+\sqrt{2015}\) nhỏ hơn 2\(\sqrt{2016}\)
hay \(\sqrt{2017}-\sqrt{2016}\) nhỏ hơn \(\sqrt{2016}-\sqrt{2015}\)