4/ $\dfrac{1}{2}x-1=\dfrac{1}{2020^2}.2020^3$
$↔\dfrac{1}{2}x-1=\dfrac{2020^3}{2020^2}=2020$
$↔\dfrac{1}{2}x=2021$
$↔x=2021:\dfrac{1}{2}=2021.2=4042$
5/ $(5x+1)^2=\dfrac{36}{49}$
\(↔\left[ \begin{array}{l}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{array} \right.\)
\(↔\left[ \begin{array}{l}5x=\dfrac{6}{7}-1\\5x=-\dfrac{6}{7}-1\end{array} \right.\)
\(↔\left[ \begin{array}{l}5x=-\dfrac 1 7\\5x=-\dfrac{13}{7}\end{array} \right.\)
\(↔\left[ \begin{array}{l}x=-\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{array} \right.\)
6/ $|x|=1,2$
\(↔\left[ \begin{array}{l}x=1,2\\x=-1,2\end{array} \right.\)
$|-x|=-(-0,25)$
$↔|-x|=0,25$
\(↔\left[ \begin{array}{l}-x=0,25\\-x=-0,25\end{array} \right.\)
\(↔\left[ \begin{array}{l}x=-0,25\\x=0,25\end{array} \right.\)
