Ta đc:\(a^2+a+1+\dfrac{1}{a}+\dfrac{1}{a^2}=0\\ \Leftrightarrow\left(a+\dfrac{1}{a}\right)^2+a+\dfrac{1}{a}-1=0\\ \Leftrightarrow\left(a+\dfrac{1}{a}+\dfrac{1}{2}\right)^2-1-\dfrac{1}{4}=0\\ \Leftrightarrow\left(a+\dfrac{1}{a}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)