$c)_{}$ $(x+1)^3-(x-1)(x^2+x+1)-2=0_{}$  

$⇔x^3+3x^2+3x+1-(x^3-1)-2=0_{}$ 

$⇔x^3+3x^2+3x+1-x^3+1-2=0_{}$ 

$⇔3x^2+3x=0_{}$ 

$⇔3x.(x+1)=0_{}$ 

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

$b)_{}$ $6x^2-(2x-3)(3x+2)=1_{}$ 

$⇔6x^2-(6x^2+4x-9x-6)=1_{}$ 

$⇔6x^2-(6x^2-5x-6)=1_{}$ 

$⇔6x^2-6x^2+5x+6=1_{}$ 

$⇔5x+6=1_{}$ 

$⇔5x=1-6_{}$ 

$⇔5x=-5_{}$ 

$⇔x=-1_{}$ 

$Vậy_{}$ $x_{}=-1$ 

$a)_{}$ $(x-2)^3-(x-3)(x^2+3x+9)+6.(x+1)^2=49_{}$ 

$⇔x^3-6x^2+12x-8-(x^3-27)+6.(x^2+2x+1)=49_{}$ 

$⇔x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49_{}$ 

$⇔24x+25=49_{}$ 

$⇔24x=49-25_{}$ 

$⇔24x=24_{}$ 

$⇔x=1_{}$ 

$Vậy_{}$ $x_{}=1$ 

a) `(x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=49`

`⇒x^3-6x^2+12x-8-x^3+27+6(x+1)^2-49=0`

`⇒24x-24=0`

`⇒x-1=0`

`⇒x=1`

b) `6x^2-(2x-3)(3x+2)=1`

`⇒6x^2-6x^2+5x+6-1=0`

`⇒5(x+1)=0`

`⇒x+1=0`

`⇒x=-1`