a,
x² - 2( x + 3 ) = x - 6
⇔ x² - 2x - 6 = x - 6
⇔ x² - 3x = 0
⇔ x ( x - 3 ) = 0
⇔ \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy x ∈ { 0 ; 3 }
b, ĐK: x ≥ - 1
|3-2x|=x+1
⇔ \(\left[ \begin{array}{l}3-2x=x+1 ( x \geq \frac{2}{3} ) \\3-2x=-x-1(x <\frac{2}{3})\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}-3x=-2\\-x=-4\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2/3\\x=4\end{array} \right.\)
Vậy....