Đáp án:
$\begin{array}{l}
b)\dfrac{{x + 1}}{4} + \dfrac{{x + 2}}{3} = \dfrac{{x - 1}}{6} + \dfrac{{x - 2}}{7}\\
\Rightarrow \dfrac{{x + 1}}{4} + 1 + \dfrac{{x + 2}}{3} + 1\\
= \dfrac{{x - 1}}{6} + 1 + \dfrac{{x - 2}}{7} + 1\\
\Rightarrow \dfrac{{x + 1 + 4}}{4} + \dfrac{{x + 2 + 3}}{3} = \dfrac{{x - 1 + 6}}{6} + \dfrac{{x - 2 + 7}}{7}\\
\Rightarrow \dfrac{{x + 5}}{4} + \dfrac{{x + 5}}{3} = \dfrac{{x + 5}}{6} + \dfrac{{x + 5}}{7}\\
\Rightarrow \left( {x + 5} \right).\left( {\dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{6} - \dfrac{1}{7}} \right) = 0\\
\Rightarrow x + 5 = 0\\
\Rightarrow x = - 5\\
\text{Vậy}\,x = - 5\\
3)\dfrac{{x - 2020}}{{204}} = \dfrac{{x - 2020}}{{2022}} = \dfrac{{x - 2020}}{{2024}}\\
\Rightarrow x - 2020 = 0\\
\Rightarrow x = 2020\\
\text{Vậy}\,x = 2020
\end{array}$
