\(\text{Cho f(x)=0}\\\Leftrightarrow x^2-x+1=0\\\Leftrightarrow x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=0\\\Leftrightarrow x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}=-\dfrac{3}{4}\\\Leftrightarrow \left(x^2-\dfrac{1}{2}x\right)-\left(\dfrac{1}{2}x-\dfrac{1}{4}\right)=-\dfrac{3}{4}\\\Leftrightarrow x\left(x-\dfrac{1}{2}\right)-\dfrac{1}{2}\left(x-\dfrac{1}{2}\right)=-\dfrac{3}{4}\\\Leftrightarrow \left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}\right)=-\dfrac{3}{4}\\\Leftrightarrow \left(x-\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\quad\text{(vô lí vì }\left(x-\dfrac{1}{2}\right)^2\ge 0\quad\forall x;-\dfrac{3}{4}<0)\\\text{Vậy đa thức f(x) không có nghiệm.}\)
