\(\left(4x-1\right)^3=27\Leftrightarrow4x-1=3\Leftrightarrow4x=4\Leftrightarrow x=1\)
\(\left(4x-1\right)^3=27\\ \Leftrightarrow\left(4x-1\right)^3=3^3\\ \Leftrightarrow4x-1=3\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)
Ta có: \(\left(4x-1\right)^3=27\)
\(\Leftrightarrow4x-1=3\)
\(\Leftrightarrow4x=4\)
hay x=1