Đáp án:
$\begin{array}{l}
a)\left| {x + \dfrac{7}{3}} \right| - \left| { - \dfrac{1}{3}} \right| = {\left( {\dfrac{{ - 1}}{3}} \right)^2}\\
\Leftrightarrow \left| {x + \dfrac{7}{3}} \right| - \dfrac{1}{3} = \dfrac{1}{9}\\
\Leftrightarrow \left| {x + \dfrac{7}{3}} \right| = \dfrac{1}{9} + \dfrac{1}{3}\\
\Leftrightarrow \left| {x + \dfrac{7}{3}} \right| = \dfrac{4}{9}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{7}{3} = \dfrac{4}{9}\\
x + \dfrac{7}{3} = - \dfrac{4}{9}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{4}{9} - \dfrac{7}{3} = \dfrac{{ - 17}}{9}\\
x = \dfrac{{ - 4}}{9} - \dfrac{7}{3} = \dfrac{{ - 25}}{9}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 17}}{9};x = - \dfrac{{25}}{9}\\
b){2^{x + 2}} - {2^x} = 96\\
\Leftrightarrow {2^x}{.2^2} - {2^x} = 96\\
\Leftrightarrow {2^x}.\left( {{2^2} - 1} \right) = 96\\
\Leftrightarrow {2^x}.3 = 96\\
\Leftrightarrow {2^x} = 32\\
\Leftrightarrow {2^x} = {2^5}\\
\Leftrightarrow x = 5\\
Vậy\,x = 5\\
c){3^{x - 1}} + {5.3^{x - 1}} = 162\\
\Leftrightarrow {3^{x - 1}}.\left( {1 + 5} \right) = 162\\
\Leftrightarrow {3^{x - 1}}.6 = 162\\
\Leftrightarrow {3^{x - 1}} = 27\\
\Leftrightarrow {3^{x - 1}} = {3^3}\\
\Leftrightarrow x - 1 = 3\\
\Leftrightarrow x = 4\\
Vậy\,x = 4\\
d)\dfrac{{x + 1}}{{2018}} + \dfrac{{x + 1}}{{2019}} + \dfrac{{x + 1}}{{2020}} + \dfrac{{x + 1}}{{2021}} = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {\dfrac{1}{{2018}} + \dfrac{1}{{2019}} + \dfrac{1}{{2020}} + \dfrac{1}{{2021}}} \right) = 0\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = - 1\\
Vậy\,x = - 1
\end{array}$
