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

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

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

\(\left(0,25\right)^8=\left[\left(0,5\right)^2\right]^8=\left(0,5\right)^{2.8}=\left(0,5\right)^{16}\)

\(\left(0,125\right)^4=\left[\left(0,5\right)^3\right]^4=\left(0,5\right)^{3.4}=\left(0,5\right)^{12}\)

\(\left(0,0635\right)^2=\left[\left(0,5\right)^3\right]^2=\left(0,5\right)^{3.2}=\left(0,5\right)^6\)

(0,25)⁸=(0,5²)⁸=(0,5)¹⁶

(0,125)⁴=(0,5³)⁴=(0,5)¹²

Bài 31 Viết các số và dưới dạng các lũy thừa của cơ số 0,5

Lời giải:

Ta có:

Giải : Ta có:

(0,25)⁸ =[(0,5)²]⁸=(0,5)¹⁶

(0,125)⁴=[(0,5)³]⁴=(0,5)¹²

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:

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

Bài 6: 

a: \(2^{27}=8^9\)

\(3^{18}=9^9\)

b: Vì \(8^9< 9^9\)

nên \(2^{27}< 3^{18}\)

\(\left(0,36\right)^8=\left[\left(0,6\right)^2\right]^8=\left(0,6\right)^{16}\)

\(\left(0,126\right)^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)

NHỚ K CHO MK NHA

(0,36)⁸=(0,6²)⁸=(0,6)^2.8=0,6¹⁶

(0,126)⁴=(0,6³)⁴=(0,6)^3.4=0,6¹²