(0,36)⁸=(0,6)¹⁶

(0,216)⁴=(0,6)¹²

0,36⁸=0,6²⁴

0,216⁴=0,6¹²

đó mk chỉ bt thế thui

chúc b hok tốt

1.

a) x : \(\left(\dfrac{3}{4}\right)^3\) =\(\left(\dfrac{3}{4}\right)^3\)

x = \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{3}{4}\right)^3\)

x = \(\dfrac{3}{4}^{3+3}\)

x = \(\dfrac{3}{4}^6\)

x = \(\dfrac{729}{4096}\)

b) \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)

x = \(\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\)

x = \(\dfrac{2}{5}^{8-5}\)

x = \(\dfrac{2}{5}^3\)

x = \(\dfrac{8}{5}\)

2.

(0,36)\(^8\) \([\left(0,6\right)^3]^8\) = (0,6)\(^{3.8}\) = ( 0,6)\(^{24}\)

( 0,216)\(^4\) = \([\left(0,6\right)^3]^4\) = (0.6)\(^{3.4}\) = ( 0,6)\(^{12}\)

\(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)

\(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\) <=> \(x=\left(\dfrac{3}{4}\right)^{2+3}\)

=> \(x=\left(\dfrac{3}{4}\right)^5\)

b, \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)

\(x=\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\Leftrightarrow x=\left(\dfrac{2}{5}\right)^{8-5}\)

=>\(x=\left(\dfrac{2}{5}\right)^3\)

bài 2 : Với bài này ta cần áp dụng quy tắc: \(\left(x^m\right)^n=x^{m.n}\)

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

\(0,216^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)

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

Ta có:

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

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

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

(0.36)⁸=((0.6)²)⁸=(0.6)¹⁶

(0.216)⁴=((0.6)³)⁴=(0.6)¹²

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

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

\(=0,5^8\)

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

(0,25)⁸=(0,5)¹⁶

(0,125)⁴=(0,5)¹²

Ta có:

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