$a$) $(2x+3)^2 - (2x+3) = 0$
$⇔ (2x+3).[(2x+3) - 1] = 0$
$⇒$ \(\left[ \begin{array}{l}2x=-3\\2x=-2\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x = \dfrac{-3}{2}\\x=-1\end{array} \right.\)
Vậy $x$ $∈$ `{-3/2;-1}`.
$b$) $(x-5)^2 = (-x+5)^2$
$⇔ (x-5)^2 = [-(x-5)]^2$
$⇔ (x-5)^2 = (5-x)^2$
$⇔ (x-5)^2 = (x-5)^2$ (Đúng với mọi $x$)
Vậy $x$ $∈$ $R$.