Tính biểu thức:
a,
b,
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1 câu trả lời 487
\[\begin{array}{l}
\frac{{{2^{19}}{{.27}^3} + {{15.4}^9}{{.9}^4}}}{{{6^9}{{.2}^{10}} + {{12}^{10}}}}\\
= \frac{{{2^{19}}.{{\left( {{3^3}} \right)}^3} + 3.5.{{\left( {{2^2}} \right)}^9}.{{\left( {{3^2}} \right)}^4}}}{{{{\left( {2.3} \right)}^9}{{.2}^{10}} + {{\left( {{2^2}.3} \right)}^{10}}}}\\
= \frac{{{2^{19}}{{.3}^9} + {3^9}{{.2}^{18}}.5}}{{{2^{19}}{{.3}^9} + {2^{20}}{{.3}^{10}}}}\\
= \frac{{{2^{18}}{{.3}^9}\left( {2 + 5} \right)}}{{{2^{19}}{{.3}^9}\left( {1 + 2.3} \right)}}\\
= \frac{7}{{2.7}}\\
= \frac{1}{2}\\
\frac{{{{\left( { - _2^1} \right)}^3} - {{\left( {\frac{3}{4}} \right)}^3}.{{\left( { - 2} \right)}^2}}}{{2.{{\left( { - 1} \right)}^5} + {{\left( {\frac{3}{4}} \right)}^2} - \frac{3}{8}}}\\
= \frac{{\frac{{ - 1}}{8} - \frac{{27}}{{64}}.4}}{{ - 2 + \frac{9}{{16}} - \frac{3}{8}}}\\
= \frac{{\frac{{ - 1}}{8} - \frac{{27}}{{16}}}}{{\frac{{ - 29}}{{16}}}}\\
= \frac{{\frac{{ - 29}}{{16}}}}{{\frac{{ - 29}}{{16}}}}\\
= 1
\end{array}\]
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