Question

\(4x^2-4x-1>0\)

Answer 1

\(4x^2-4x-1>0\Leftrightarrow4x^2-4x+1-2>0\Leftrightarrow\left(2x-1\right)^2-2>0\)

\(\Leftrightarrow\left(2x-1\right)^2>2\Leftrightarrow\left\{{}\begin{matrix}2x-1>\sqrt{2}\\hoặc\\2x-1< -\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x>1+\sqrt{2}\\hoặc\\2x< 1-\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1+\sqrt{2}}{2}\\hoặc\\a< \dfrac{1-\sqrt{2}}{2}\end{matrix}\right.\) vậy \(x>\dfrac{1+\sqrt{2}}{2}\) hoặc \(x< \dfrac{1-\sqrt{2}}{2}\)