Đáp án:
\(\begin{array}{l}
a,\,\,\,\,\dfrac{1}{{32}}\\
b,\,\,\,\,\dfrac{1}{9}\\
c,\,\,\,\,\dfrac{{81}}{2}\\
d,\,\,\,\,\dfrac{{15}}{2}\\
e,\,\,\,\,1\\
f,\,\,\,\,64
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\dfrac{{{{18}^5}}}{{{4^5}{{.9}^5}}} = \dfrac{{{{\left( {2.9} \right)}^5}}}{{{{\left( {{2^2}} \right)}^5}{{.9}^5}}} = \dfrac{{{2^5}{{.9}^5}}}{{{2^{2.5}}{{.9}^5}}} = \dfrac{{{2^5}{{.9}^5}}}{{{2^{10}}{{.9}^5}}} = \dfrac{1}{{{2^5}}} = \dfrac{1}{{32}}\\
b,\\
\dfrac{{{{16}^2}}}{{{3^2}{{.4}^4}}} = \dfrac{{{{16}^2}}}{{{3^2}.{{\left( {{4^2}} \right)}^2}}} = \dfrac{{{{16}^2}}}{{{3^2}{{.16}^2}}} = \dfrac{1}{{{3^2}}} = \dfrac{1}{9}\\
c,\\
\dfrac{{{8^5}{{.12}^4}}}{{{4^3}{{.8}^6}}} = \dfrac{{{8^5}}}{{{8^6}}}.\dfrac{{{{12}^4}}}{{{4^3}}} = \dfrac{1}{8}.\dfrac{{{{\left( {3.4} \right)}^4}}}{{{4^3}}} = \dfrac{1}{8}.\dfrac{{{3^4}{{.4}^4}}}{{{4^3}}} = \dfrac{1}{8}{.3^4}.4\\
= \dfrac{1}{2}{.3^4} = \dfrac{{81}}{2}\\
d,\\
\dfrac{{{8^2} + {{3.4}^2} + 128}}{{{2^5}}} = \dfrac{{{{\left( {{2^3}} \right)}^2} + 3.{{\left( {{2^2}} \right)}^2} + {2^7}}}{{{2^5}}}\\
= \dfrac{{{2^6} + {{3.2}^4} + {2^7}}}{{{2^5}}} = \dfrac{{{2^4}.\left( {{2^2} + 3 + {2^3}} \right)}}{{{2^5}}} = \dfrac{{{2^4}.15}}{{{2^5}}} = \dfrac{{15}}{2}\\
e,\\
{\left( { - 1} \right)^{20}} + 1,69.{\left( {3,5 - 2,2} \right)^2} - {\left( { - 1,69} \right)^2}\\
= 1 + 1,69.1,{3^2} - 1,{69^2}\\
= 1 + 1,69.1,69 - 1,{69^2}\\
= 1 + 1,{69^2} - 1,{69^2}\\
= 1\\
f,\\
\dfrac{{{4^{19}} + {8^7}}}{{{{256}^4} + {{32}^3}}} = \dfrac{{{{\left( {{2^2}} \right)}^{19}} + {{\left( {{2^3}} \right)}^7}}}{{{{\left( {{2^8}} \right)}^4} + {{\left( {{2^5}} \right)}^3}}} = \dfrac{{{2^{38}} + {2^{21}}}}{{{2^{32}} + {2^{15}}}}\\
= \dfrac{{{2^{21}}.\left( {{2^{17}} + 1} \right)}}{{{2^{15}}.\left( {{2^{17}} + 1} \right)}} = \dfrac{{{2^{21}}}}{{{2^{15}}}} = {2^6} = 64
\end{array}\)
