Giải phương trình 6x^3 + x^2 =2x (x+2)^2 – 2x(2x+3) = (x+1)^2

Giải thích các bước giải:

$a) 6x³ +x² = 2x$

`⇔ 6x^3 +x^2 -2x = 0`
`⇔ 6x².(x +2/3) -3x.(x +2/3) = 0`

`⇔ (x +2/3).(6x² -3x) = 0`

`⇔ 3x.(x +2/3).(2x -1) = 0`

$⇔ \left[ \begin{array}{l}3x=0\\x +\dfrac{2}{3}=0\\2x -1 = 0\end{array} \right. ⇔ \left[ \begin{array}{l}x=0\\x=\dfrac{-2}{3}\\x = \dfrac{1}{2}\end{array} \right.$

Vậy `S = {0; -2/3; 1/2}`

$b) (x +2)² -2x.(2x +3) = (x +1)²$

`⇔ (x +2)² -(x +1)² -2x.(2x +3) = 0`

`⇔ (x +2 -x -1).(x +2 +x +1) -4x² -6x = 0`

`⇔ 2x +3 -4x² -6x = 0`

`⇔ -4x² -4x +3 = 0`

`⇔ -4x.(x -1/2) -6.(x -1/2) = 0`

`⇔ (x -1/2).(-4x -6) = 0`

$⇔ \left[ \begin{array}{l}x -\dfrac{1}{2}=0\\-4x -6=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{-3}{2}\end{array} \right.$

Vậy `S = {1/2; -3/2}`