Đáp án:
$\begin{array}{l}
a){25.5^3}.\dfrac{1}{{625}}{.5^3}\\
= {5^2}{.5^3}{.5^3}.\dfrac{1}{{{5^4}}}\\
= {5^4}\\
b){4^2}.32:{2^3}\\
= {2^{2.2}}{.2^5}:{2^3}\\
= {2^{4 + 5 - 3}}\\
= {2^6}\\
c){27.5^3}{.3^3}{.32^{ - 1}}:125\\
= {3^3}{.3^3}{.5^3}.\dfrac{1}{{{2^5}{{.5}^3}}}\\
= \dfrac{{{3^6}}}{{{2^5}}}\\
d){5^6}{.20^{ - 1}}{.2^2}.{\left( { - 3} \right)^2}:125\\
= {5^6}.\dfrac{1}{{{2^2}.5}}{.2^2}{.3^2}.\dfrac{1}{{{5^3}}}\\
= {5^2}{.3^2}\\
e){27.3^3}.\dfrac{1}{{81}}{.3^{27}}\\
= {3^3}{.3^3}{.3^{27}}.\dfrac{1}{{{3^4}}}\\
= {3^{29}}\\
f){5^3}.625:{5^2}\\
= {5^3}{.5^4}.\dfrac{1}{{{5^2}}}\\
= {5^5}\\
g){64.125.3^3}.\dfrac{1}{{27}}:{25^3}\\
= {2^6}{.5^3}{.3^3}.\dfrac{1}{{{3^3}{{.5}^6}}}\\
= \dfrac{{{2^6}}}{{{5^3}}}\\
h){\left( {\dfrac{1}{7}} \right)^2}{.7^{ - 1}}:\dfrac{1}{{{{49}^2}}}\\
= \dfrac{1}{{{7^2}.7}}{.7^{2.2}}\\
= 7
\end{array}$
