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Đáp án:$\begin{array}{l}
A = - 2020\\
H = - \dfrac{3}{2}
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
A = 2020.\dfrac{{0,6 - \dfrac{1}{3} + \dfrac{3}{{11}}}}{{1,4 - \dfrac{7}{9} + \dfrac{7}{{11}}}}.\dfrac{{ - 1\dfrac{1}{6} + 0,875 - 0,7}}{{0,5 - \dfrac{3}{8} + \dfrac{3}{{10}}}}\\
= 2020.\dfrac{{\dfrac{3}{5} - \dfrac{1}{3} + \dfrac{3}{{11}}}}{{\dfrac{7}{5} - \dfrac{7}{9} + \dfrac{7}{{11}}}}.\dfrac{{\dfrac{{ - 7}}{6} + \dfrac{7}{8} - \dfrac{7}{{10}}}}{{\dfrac{1}{2} - \dfrac{3}{8} + \dfrac{3}{{10}}}}\\
= 2020.\dfrac{{3.\left( {\dfrac{1}{5} - \dfrac{1}{9} + \dfrac{1}{{11}}} \right)}}{{7.\left( {\dfrac{1}{5} - \dfrac{1}{9} + \dfrac{1}{{11}}} \right)}}.\dfrac{{ - 7.\left( {\dfrac{1}{6} - \dfrac{1}{8} + \dfrac{1}{{10}}} \right)}}{{3.\left( {\dfrac{1}{6} - \dfrac{1}{8} + \dfrac{1}{{10}}} \right)}}\\
= 2020.\dfrac{3}{7}.\dfrac{{ - 7}}{3}\\
= - 2020\\
H = \dfrac{{{2^{19}}{{.27}^3}.5 - 15.{{\left( { - 4} \right)}^9}{{.9}^4}}}{{{6^9}{{.2}^{10}} - {{\left( { - 12} \right)}^{10}}}}\\
= \dfrac{{{2^{19}}.{{\left( {{3^3}} \right)}^3}.5 + 3.5.{{\left( {{2^2}} \right)}^9}.{{\left( {{3^2}} \right)}^4}}}{{{2^9}{{.3}^9}{{.2}^{10}} - {{\left( {{2^2}.3} \right)}^{10}}}}\\
= \dfrac{{{2^{19}}{{.3}^9}.5 + {{3.5.2}^{18}}{{.3}^8}}}{{{2^{19}}{{.3}^9} - {2^{20}}{{.3}^{10}}}}\\
= \dfrac{{{2^{18}}{{.3}^9}.5\left( {2 + 1} \right)}}{{{2^{19}}{{.3}^9}.\left( {1 - 2.3} \right)}}\\
= \dfrac{{5.3}}{{2.\left( { - 5} \right)}}\\
= - \dfrac{3}{2}
\end{array}$
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