tìm x, 2020^(x-1).(2.x+3)=1 (5.x+1)^2=36/49 |x-2|-|1-2.x|=0 chiều mình nộp rùi help me gạch chéo là phân số

$a)$ $2020^{x – 1}$ . $(2x + 3) = 1$

$$⇒ \left \{ {{2020^{x – 1} = 1} \atop {2x + 3 = 1}} \right.$$

$$⇒ \left \{ {{x = 1} \atop {x = -1}} \right.$$

$b) (5x + 1)^{2} = \dfrac{36}{49}$

$⇒  (5x + 1)^{2} = (\dfrac{6}{7})^{2}$ 

$⇒ 5x + 1           = \dfrac{6}{7}$

$⇒ \left[ \begin{array}{l}5x + 1 = \dfrac{6}{7}\\5x + 1 = \dfrac{-6}{7}\end{array} \right.$

$⇒ \left[ \begin{array}{l}x = \dfrac{-1}{35}\\x = \dfrac{-13}{35}\end{array} \right.$

$c) | x – 2 | – | 1 – 2x | = 0$

Ta thấy $| x – 2 | ≥ 0 ; | 1 – 2x | ≥ 0$

$⇒ x – 2 = 0$

$⇒ x       = 2$

hoặc $1 – 2x = 0 ⇒ x = \dfrac{1}{2}$

Vậy $x ∈ ∅$