$a)$ $2020^{x - 1}$ . $(2x + 3) = 1$
$$⇒ \left \{ {{2020^{x - 1} = 1} \atop {2x + 3 = 1}} \right.$$
$$⇒ \left \{ {{x = 1} \atop {x = -1}} \right.$$
$b) (5x + 1)^{2} = \dfrac{36}{49}$
$⇒ (5x + 1)^{2} = (\dfrac{6}{7})^{2}$
$⇒ 5x + 1 = \dfrac{6}{7}$
$⇒ \left[ \begin{array}{l}5x + 1 = \dfrac{6}{7}\\5x + 1 = \dfrac{-6}{7}\end{array} \right.$
$⇒ \left[ \begin{array}{l}x = \dfrac{-1}{35}\\x = \dfrac{-13}{35}\end{array} \right.$
$c) | x - 2 | - | 1 - 2x | = 0$
Ta thấy $| x - 2 | ≥ 0 ; | 1 - 2x | ≥ 0$
$⇒ x - 2 = 0$
$⇒ x = 2$
hoặc $1 - 2x = 0 ⇒ x = \dfrac{1}{2}$
Vậy $x ∈ ∅$
