helppp Giải phương trình (2x-1)(x-1)(3x+1)=1

`(2x-1)(x-1)(3x+1)=1`

`<=>6x^3+2x^2-6x^2-2x-3x^2-x+3x+1=1`

`<=>6x^3-7x^2=0`

`<=>x^2(6x-7)=0`

`<=>[(x^2=0),(6x-7=0):}`

`<=>[(x=0),(6x=7):}`

`<=>`$\left[\begin{matrix} x=0\\ x=\dfrac{7}{6}\end{matrix}\right.$

Vậy `S={0; 7/6}`

$(2x-1)(x-1)(3x+1)=1$

$\Leftrightarrow (2x^2-2x-x+1)(3x+1)=1$

$\Leftrightarrow (2x^2-3x+1)(3x+1)=1$

$\Leftrightarrow 6x^3+2x^2-9x^2-3x+3x+1=1$

$\Leftrightarrow 6x^3-7x^2=0$

$\Leftrightarrow x^2(6x-7)=0$

$\Leftrightarrow \left[ \begin{array}{l}x^2=0\\6x-7=0\end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l}x=0\\x=\dfrac{7}{6}\end{array} \right.$

Vậy $S=\left\{\dfrac{7}{6};0\right\}$