Question

simply the following ​

Answer 1

Answer:

Answer for the problems are given in the attachments.

Step-by-step explanation:

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Answer 2

For the given questions we need to apply laws of exponents.

(i)

First law we will be applying is ⇒ $\sf (a^m)^n=a^{m\times n}$

$\sf \{(\frac{2}{3})^2\}^3\times(\frac{1}{3} )^{-4}\times (3)^{-2}\times (2)^{-1}$

$\sf (\frac{2}{3})^{2\times 3}\times(\frac{1}{3} )^{-4}\times (3)^{-2}\times (2)^{-1}$

$\sf (\frac{2}{3})^{6}\times(\frac{1}{3} )^{-4}\times (3)^{-2}\times (2)^{-1}$

Now we will apply this law $\sf a^{-m}=\frac{1}{a^m}$

$\sf (\frac{2}{3})^{6}\times(\frac{1}{3} )^{-4}\times (3)^{-2}\times (2)^{-1}$

$\sf (\frac{2}{3})^{6}\times(3 )^{4}\times (\frac{1}{3} )^{2}\times (\frac{1}{2} )^{1}$

Now we have to give the values and solve.

$\sf (\frac{2}{3})^{6}\times(3 )^{4}\times (\frac{1}{3} )^{2}\times (\frac{1}{2} )^{1}$

$\sf \frac{64}{729}\times81\times \frac{1}{9} \times \frac{1}{2}$

$\sf \frac{64}{729}\times81\times \frac{1}{18}$

$\sf \frac{64}{9} \times \frac{1}{18}$

$\sf \frac{64}{9} \times \frac{1}{18}$

$\sf \frac{64}{162}$

$\frac{32}{81}$

∴$\sf \{(\frac{2}{3})^2\}^3\times(\frac{1}{3} )^{-4}\times (3)^{-2}\times (2)^{-1}=\frac{32}{81}$

(ii)

First law we will be applying would be ⇒ $\sf a^{-m}=\frac{1}{a^m}$

$\sf (4^{-1}\div6^{-1})^3$

$\sf (\frac{1}{4}\div \frac{1}{6})^3$

$\sf (\frac{1}{4} \times 6)^3$

$\sf (\frac{1}{2}\times 3)^3$

$\sf (\frac{3}{2})^3$

Now we have to give the values to find the answer.

$\sf (\frac{3}{2})^3$

$\sf \frac{27}{8}$

∴ $\sf (4^{-1}\div6^{-1})^3 =\frac{27}{8}$