`(2x - 1)^3 = 4(2x - 1)`
`=>8x^3 - 12x^2 + 6x - 1 = 8x - 4`
`=> 8x^3 - 12x^2 + 6x - 8x -1 + 4= 0`
`=> 8x^3 - 12x^2 -2x + 3 =0`
`=> 4x^2(2x - 3) - (2x - 3) =0`
`=> (2x - 3)(4x^2 - 1) = 0`
TH1:
`2x - 3 = 0`
`=> 2x = 0+3`
`=> 2x = 3`
`=> x = 3:2`
`=> x = 3/2`
TH2:
`4x^2 -1 = 0`
`=> 4x^2 = 0+1`
`=> 4x^2 = 1`
`=> x^2 = 1/4`
`=> x = 1/2` hoặc `x = -1/2`
Vậy `x∈{3/2 ; 1/2 ; -1/2}`
(2\(x\) - 1)3 = 4.(2\(x\) - 1)
(2\(x\) - 1)3 - 4.(2\(x\) - 1) = 0
(2\(x\) - 1).[(2\(x\) - 1)2 - 4] = 0
\(\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=1\\\left(2x-1\right)^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x-1=-2\\2x-1=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x=-2+1\\2x=2+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x=-1\\2x=3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- \(\dfrac{1}{2}\); \(\dfrac{1}{2}\); \(\dfrac{3}{2}\)}
