d) Ta có: \(5x\left(x-2021\right)-x+2021=0\)
\(\Leftrightarrow\left(x-2021\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{5}\end{matrix}\right.\)
e: Ta có: \(9x^2-\dfrac{1}{4}=0\)
\(\Leftrightarrow9x^2=\dfrac{1}{4}\)
\(\Leftrightarrow x^2=\dfrac{1}{36}\)
hay \(x\in\left\{\dfrac{1}{6};-\dfrac{1}{6}\right\}\)