Đáp án:
c. -1
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{3}{4}:\dfrac{1}{9} + \dfrac{9}{4}\\
= \dfrac{3}{4}.9 + \dfrac{9}{4} = \dfrac{{27}}{4} + \dfrac{9}{4}\\
= 8\\
b.\dfrac{{45}}{{19}} - {\left( {\dfrac{1}{2} + {{\left( {\dfrac{1}{3} + 4} \right)}^{ - 1}}} \right)^{ - 1}}\\
= \dfrac{{45}}{{19}} - {\left( {\dfrac{1}{2} + {{\left( {\dfrac{{13}}{3}} \right)}^{ - 1}}} \right)^{ - 1}}\\
= \dfrac{{45}}{{19}} - {\left( {\dfrac{1}{2} + \dfrac{3}{{13}}} \right)^{ - 1}}\\
= \dfrac{{45}}{{19}} - {\left( {\dfrac{{19}}{{26}}} \right)^{ - 1}}\\
= \dfrac{{45}}{{19}} - \dfrac{{26}}{{19}} = 1\\
c.\dfrac{{{{5.2}^{2.15}}{{.3}^{2.9}} - {2^2}{{.3}^{20}}{{.2}^{3.9}}}}{{{{5.2}^{10}}{{.2}^{19}}{{.3}^{19}} - {{7.2}^{29}}{{.3}^{3.6}}}}\\
= \dfrac{{{{5.2}^{30}}{{.3}^{18}} - {2^{29}}{{.3}^{20}}}}{{{{5.2}^{29}}{{.3}^{19}} - {{7.2}^{29}}{{.3}^{18}}}}\\
= \dfrac{{{2^{29}}{{.3}^{18}}\left( {5.2 - {{2.3}^2}} \right)}}{{{2^{29}}{{.3}^{18}}\left( {5.3 - 7} \right)}}\\
= \dfrac{{ - 8}}{8} = - 1
\end{array}\)
