Lời giải:
$4x^2-x-1=0$
$\Leftrightarrow [(2x)^2-2.2x.\frac{1}{4}+(\frac{1}{4})^2]-\frac{17}{16}=0$
$\Leftrightarrow (2x-\frac{1}{4})^2=\frac{17}{16}$
$\Rightarrow 2x-\frac{1}{4}=\frac{\sqrt{17}}{4}$ hoặc $2x-\frac{1}{4}=\frac{-\sqrt{17}}{4}$
$\Rightarrow x=\frac{1+\sqrt{17}}{8}$ hoặc $x=\frac{1-\sqrt{17}}{8}$


