\(a=\dfrac{\dfrac{2+\sqrt{3}}{2}}{1+\sqrt{\dfrac{2+\sqrt{3}}{2}}}+\dfrac{\dfrac{2-\sqrt{3}}{2}}{1-\sqrt{\dfrac{2-\sqrt{3}}{2}}}=\dfrac{\dfrac{2+\sqrt{3}}{2}}{1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}}+\dfrac{\dfrac{2-\sqrt{3}}{2}}{1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}}\)
\(a=\dfrac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(a=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}=\dfrac{6}{6}=1\)
Thay \(a=1\) vào pt: \(2013.1^2-2014.1+1=2013-2014+1=0\)
\(\Rightarrow a\) là một nghiệm của \(2013x^2-2014x+1=0\)