c)Ta có: \(\left(\sqrt{2015}+\sqrt{2017}\right)^2=2017+2\sqrt{\left(2016-1\right)\left(2016+1\right)}+2015\\ =2.2016+2\sqrt{2016^2-1}\)
\(\left(2\sqrt{2016}\right)^2=4.2016=2.2016+2.\sqrt{2016^2}\)
Vì \(2\sqrt{2016^2-1}< 2\sqrt{2016^2}\) nên \(\left(\sqrt{2015}+\sqrt{2017}\right)^2< \left(2\sqrt{2016}\right)^2\)
\(\Rightarrow\sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}\)
a) Ta có \(\sqrt{5}+\sqrt{7}>\sqrt{4}+\sqrt{4}=2+2=4=\sqrt{16}>\sqrt{13}\)
Vậy \(\sqrt{5}+\sqrt{7}>\sqrt{13}\)
b) Ta có \(\sqrt{15}.\sqrt{17}=\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\)
\(16=\sqrt{16^2}\)
Vì \(16^2>16^2-1\) nên \(\sqrt{16^2}>\sqrt{16^2-1}\Leftrightarrow16>\sqrt{15}.\sqrt{17}\)
\(\text{c) Ta có }:\left(\sqrt{2015}+\sqrt{2017}\right)^2=2017+2\sqrt{2015\cdot2017}+2015\\ =2016+1+2\sqrt{\left(2016-1\right)\left(2016+1\right)}+2016-1\\ =2\cdot2016+2\sqrt{2016^2-1}\\ >2\cdot2016+2\sqrt{2016^2}\\ =2\cdot2016+2\cdot2016\\ =4\cdot2016=\left(2\sqrt{2016}\right)^2\)
\(\Rightarrow\sqrt{2015}+\sqrt{2017}>2\sqrt{2016}\)
Vậy........
