Ta có:

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

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

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

\(=0,5^8\)

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

\(\begin{array}{l}{\left( {0,25} \right)^8} = {\left[ {{{\left( {0,5} \right)}^2}} \right]^8}=(0,5)^{2.8} = {\left( {0,5} \right)^{16}};\\{\left( {0,125} \right)^4} = {\left[ {{{\left( {0,5} \right)}^3}} \right]^4} =(0,5)^{3.4}= {\left( {0,5} \right)^{12}};\\{\left( {0,0625} \right)^2} = {\left[ {{{\left( {0,5} \right)}^4}} \right]^2} =(0,5)^{4.2}= {\left( {0,5} \right)^8}\end{array}\)

\(2S=2^3+2^3+2^4+...+2^{21}\\ S=2^{21}+2^3-2^2-2^2=2^{21}+8-4-4=2^{21}\)

M=2^2+2^2+2^3+...+2^2022

=>2M=2^3+2^3+2^4+...+2^2022+2^2023

=>2M-M=2^2023+2^2022+...+2^4+2^3+2^3-2^2022-...-2^3-2^2-2^2

=>M=2^2023+2^3-2^2-2^2

=>M=2^2023

Ta có:

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

a)\(4^3.2^4\div\left(4^2.\frac{1}{32}\right)\)

\(=\left(2^2\right)^3.2^4\div\left(2^2\right)^2\div32\).

\(=2^{\left(2.3\right)}.2^4\div2^{\left(2.2\right)}\div2^5\)

\(=2^6.2^4\div2^4\div2^5\)

\(=2^{6+4-4-5}=2^1\)

b)\(\left(\frac{1}{5}\right)^5=\frac{1}{5^5}=\left|5^5\right|=5^{-5}\)

\(\frac{1}{125}=\frac{1}{5^3}=\left|5^3\right|=5^{-3}\)

c)\(\frac{4}{25}=\frac{2^2}{5^2}=\left(\frac{2}{5}\right)^2=0,4^2\)

\(\frac{-8}{125}=\frac{-2^3}{5^3}=\left(\frac{-2}{5}\right)^2=-0,4^3=0,4^{-3}\)

\(\frac{16}{625}=\frac{2^4}{5^4}=\left(\frac{2}{5}\right)^4=0,4^4\)

a) 8 = 2³

42⁵ = 2⁵.3⁵.7⁵

16 = 2⁴

b) (0,09)³ = (3/10)⁶

(3/10)⁸ = (3/10)⁸

0,027 = (3/10)³

`@` `\text {Ans}`

`\downarrow`

`a)`

`8 = 2^3`

`32^5` chứ ạ?

`32^5 = (2^5)^5 = 2^10`

`16 = 2^4`

`b)`

`(0,09)^3 = (0,3^2)^3 = 0,3^6` hay `(3/10)^6`

`(3/10)^8 = (3/10)^8`

`(0,027) = (0,3)^3` hay `(3/10)^3`

`@` `\text {Kaizuu lv uuu}`