Tìm x .

`a)` `6x ( 5x - 2 ) - ( 5x - 2 ) . 2 = 0`

`=> ( 5x - 2 ) ( 6x - 1.2 ) = 0`

`=> ( 5x - 2 ) ( 6x - 2 ) = 0`

`=>` $\left[\begin{matrix} 5x-2=0\\ 6x-2=0\end{matrix}\right.$

`=>` $\left[\begin{matrix} 5x=2\\ 6x=2\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=\dfrac{2}{5}\\ x=\dfrac{2}{6}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=\dfrac{2}{5}\\ x=\dfrac{1}{3}\end{matrix}\right.$

Vậy `x ∈ { 2/5 ; 1/3 }`

`c)` `8x ( x - 2017 ) - 2x + 4034 = 0`

`=> 8x ( x - 2017 ) - ( 2x - 4034 ) = 0`

`=> 8x ( x - 2017 ) - 2 ( x - 2017 ) = 0`

`=> ( x - 2017 ) ( 8x - 2 ) = 0`

`=>` $\left[\begin{matrix} x-2017=0\\ 8x-2=0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=2017\\ 8x=2\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=2017\\ x=\dfrac{2}{8}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=2017\\ x=\dfrac{1}{4}\end{matrix}\right.$

Vậy `x ∈ { 2017 ; 1/4 }`

`e)` `x^4 + 5x^3 - 8x - 40 = 0 `

`=> ( x^4 + 5x^3 ) - ( 8x + 40 ) = 0`

`=> x^3 ( x + 5 ) - 8 ( x + 5 ) = 0`

`=> ( x + 5 ) ( x^3 - 8 ) = 0`

`=> ( x + 5 ) ( x - 2 ) ( x^2 + 2x + 4 ) = 0`

`=>` $\left[\begin{matrix} x+5=0\\x-2=0\\ x^2+2x+4=0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=-5\\x=2\\ x^2+2x+4=0 ( \text{loại})\end{matrix}\right.$

Vậy `x ∈ { -5 ; 2 }`