Đáp án:
$\begin{array}{l}
\left| {{x^2} + 1} \right| - \left| {x + 1} \right| = {x^2} - 2x - 2018\\
\Rightarrow {x^2} + 1 - \left| {x + 1} \right| = {x^2} - 2x - 2018\left( {do:{x^2} + 1 \ge 1 > 0\forall x} \right)\\
\Rightarrow \left| {x + 1} \right| = 2x + 2019\\
\Rightarrow \left[ \begin{array}{l}
x + 1 = 2x + 2019\\
x + 1 = - 2x - 2019
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = - 2018\\
x = \frac{{ - 2020}}{3}
\end{array} \right.\\
H = \frac{{{2^{19}}{{.27}^3}.5 - 15.{{\left( { - 4} \right)}^9}{{.9}^4}}}{{{6^9}{{.2}^{10}} - {{\left( { - 12} \right)}^{10}}}}\\
= \frac{{{2^{19}}{{.3}^{3.3}}.5 + {{3.5.2}^{2.9}}{{.3}^{2.4}}}}{{{2^9}{{.3}^9}{{.2}^{10}} - {3^{10}}{{.4}^{10}}}}\\
= \frac{{5\left( {{2^{19}}{{.3}^9} + {2^{18}}{{.3}^9}} \right)}}{{{2^{19}}{{.3}^9} - {3^{10}}{{.2}^{20}}}}\\
= \frac{{{{5.2}^{18}}{{.3}^9}\left( {2 + 1} \right)}}{{{2^{19}}{{.3}^9}\left( {3 - 2} \right)}}\\
= \frac{{5.3}}{{2.1}} = \frac{{15}}{2}
\end{array}$
