Ta có:

\(\begin{array}{l}{\left( {\frac{1}{9}} \right)^5} = {[{\left( {\frac{1}{3}} \right)^2}]^5} = {(\frac{1}{3})^{2.5}} = {(\frac{1}{3})^{10}};\\{\left( {\frac{1}{{27}}} \right)^7} = {[{(\frac{1}{3})^3}]^7} = {(\frac{1}{3})^{3.7}} = {(\frac{1}{3})^{21}}\end{array}\)

Ta có:

\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)

Ta có: +) \({({2^2})^3} = {2^2}{.2^2}{.2^2} = {2^{2 + 2 + 2}} = {2^6}\)

+) \({\left[ {{{( - 3)}^2}} \right]^2} = {( - 3)^2}.{( - 3)^2} = {( - 3)^{2 + 2}} = {( - 3)^4}\)

\(\left(\dfrac{1}{27}\right)^5\) = \(\left(\dfrac{1}{3^3}\right)^5\) = \(\left(\dfrac{1}{3}\right)^{15}\)

\(\left(\dfrac{1}{27}\right)^5=\left[\left(\dfrac{1}{3}\right)^3\right]^5=\dfrac{1}{3}^{3.5}=\dfrac{1}{3}^{15}\)

\(\left(\dfrac{1}{27}\right)^5=\left(\dfrac{1}{3^3}\right)^5=\left(3^{-3}\right)^5=3^{-3.5}=3^{-15}\)

Đề hơi sai nên mình sửa lại xíu :>>

Phần b mình lười viết dưới dạng phân số nên bạn viết dưới dạng phân số giùm mình :>>

a) 8² . 32⁴ = ( 2³ )² . ( 2⁵ )⁴

= 2⁸ . 2²⁰ = 2²⁸

b) 8¹⁵ : 9⁷ = ( 3⁴ )⁵ : ( 3² )⁷ 

= 3²⁰ : 3²¹ = \(\frac{1}{3}\)

\(\left(0,0625\right)^2=\left(0,5^4\right)^2\)

\(=0,5^{4\cdot2}\)

\(=0,5^8\)

\(\left(0,0625\right)^2=\left[\left(0,5\right)^4\right]^2=\left(0,5\right)^8\)

Sai rồi bạn, phải là

a)5⁶=5³.5³

b)7⁸=7⁴.7⁴

c)8⁵=8².8³

\(a,5^6=15625\)

\(b,7^8=5764801\)

\(c,8^5=32768\)

tks  2 bạn :D