Đáp án:

e. -1

Giải thích các bước giải:

\(\begin{array}{l}
a.{\left( {\dfrac{{4.8}}{{{9^2}}}} \right)^2}.{\left( {\dfrac{9}{8}} \right)^5}{.4^3}\\
 = \dfrac{{{4^2}{{.8}^2}{{.9}^5}{{.4}^3}}}{{{9^4}{{.8}^5}}}\\
 = \dfrac{{{4^5}.9}}{{{8^3}}}\\
 = \dfrac{{{4^5}.9}}{{{2^3}{{.4}^3}}}\\
 = \dfrac{{{4^2}.9}}{{{2^3}}}\\
 = \dfrac{{{2^4}.9}}{{{2^3}}}\\
 = 2.9 = 18\\
b. - \dfrac{{{2^3}{{.3}^4}}}{{{3^3}}} + \dfrac{{{3^2}{{.2}^5}}}{{{2^4}}}\\
 =  - {2^3}.3 + {3^2}.2\\
 = 2.3\left( { - {2^2} + 3} \right)\\
 = 6.\left( { - 1} \right)\\
 =  - 6\\
c. - {5^{10}}:{5^6}:{\left( { - 5} \right)^3}\\
 = {5^{10}}:{5^6}:{5^3}\\
 = {5^4}:{5^3} = 5\\
d.\dfrac{{{2^9}}}{{{3^9}}}.\dfrac{{{3^8}}}{{{2^8}}}\\
 = \dfrac{2}{3}\\
e{.5^2}.\dfrac{{\left( { - 1} \right)}}{{{5^3}}} + \dfrac{1}{5} - 2.\dfrac{1}{{{2^2}}} - \dfrac{1}{2}\\
 =  - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{2} - \dfrac{1}{2}\\
 =  - 1
\end{array}\)