Đáp án:

$1)
a) 27\\
b) \dfrac{512}{23}\\
c)
\dfrac{-4}{7}$

Giải thích các bước giải:

$1)
a) 3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^2}\\
=\dfrac{3^2}{3^5}.\dfrac{(3^4)^2}{3^2}\\
=\dfrac{1}{3^3}.\dfrac{3^8}{3^2}\\
=\dfrac{3^8}{3^5}\\
=3^3\\
=27\\
b) (4.2^5).(2^3.\dfrac{1}{46}\\
=2^2.2^5\dfrac{2^3}{2.23}\\
=\dfrac{2^7.4}{23}\\
=\dfrac{512}{23}\\
c)
\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\\
=\dfrac{(2^2)^6.(3^2)^5+(2.3)^9.2^3.3.5}{-(2^3)^4.3^{12}-(2.3)^{11}}\\
=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\\
=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\\
=\dfrac{2^{12}.3^{10}(1+5)}{-2^{11}.3^{11}(2.3+1)}\\
=\dfrac{2.6}{-3.7}\\
=\dfrac{-4}{7}\\
2)
5^5-5^4+5^3\\
=5^3(5^2-5+1)\\
=5^3.21$
Do $21\vdots 7\Rightarrow 21.5^3\vdots 7$
$\Rightarrow 5^5-5^4+5^3\vdots 7$