Sửa:
\(\sqrt{4x^2-2x+\dfrac{1}{4}}=4x^3-x^2-8x+2\)
ĐKXĐ: \(x\in R\)
\(\Leftrightarrow\sqrt{4\left(x-\dfrac{1}{4}\right)^2}=4x\left(x^2-2\right)-\left(x^2-2\right)\)
\(\Leftrightarrow2\left(x-\dfrac{1}{4}\right)=\left(x^2-2\right)\left(4x-1\right)\\ \Leftrightarrow\left(x^2-2\right)\left(4x-1\right)-\dfrac{1}{2}\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x^2-2-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x-\sqrt{\dfrac{5}{2}}\right)\left(x+\sqrt{\dfrac{5}{2}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\sqrt{\dfrac{5}{2}}\\x=-\sqrt{\dfrac{5}{2}}\end{matrix}\right.\)
Vậy..................