\(\sqrt{1+2016^2+\dfrac{2016^2}{2017^2}}+\dfrac{2016}{2017}\)(1)
Đặt x=2016
(1)\(\Leftrightarrow\)\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}\)+\(\dfrac{2016}{2017}\)
\(\Leftrightarrow\)\(\sqrt{\dfrac{\left(a+1\right)^2+\left(a+a^2\right)^2+a^2}{\left(a+1\right)^2}}\)+\(\dfrac{2016}{2017}\)(2)
Xét:\(\left(a+1\right)^2+\left(a^2+a\right)^2+a^2\)\(=\)\(\left(a^2+a\right)^2+a^2+2a+1+a^2=\left(a^2+a\right)^2+1+2\left(a^2+a\right)=\left(a^2+a+1\right)^2\)
(2)\(\Leftrightarrow\)\(\sqrt{\dfrac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}\)+\(\dfrac{2016}{2017}\)=\(\dfrac{a^2+a+1}{a+1}+\dfrac{2016}{2017}=\dfrac{a^2+a+1}{a+1}+\dfrac{a}{a+1}=\dfrac{a^2+2a+1}{a+1}=\dfrac{\left(a+1\right)^2}{a+1}=a+1=2017\)
