Giải thích các bước giải:

$(\dfrac13)^{12}\cdot (\dfrac1{27})^{-5}+(0,4)^{-4}\cdot 25^{-2}\cdot (\dfrac1{32})^{-1}$

$=\dfrac1{3^{12}}\cdot (\dfrac1{3^3})^{-5}+(\dfrac25)^{-4}\cdot (5^2)^{-2}\cdot 32$

$=\dfrac1{3^{12}}\cdot \dfrac1{3^{-15}}+\dfrac{2^{-4}}{5^{-4}}\cdot 5^{-4}\cdot 2^5$

$=\dfrac1{3^{-3}}+ 2$

$=3^3+2$

$=27+2$

$=29$