How do you simplify the given division $ \dfrac{\sqrt{64}}{\sqrt{81}} $ ?

Hint: We start solving the problem by equating the given division to a variable. We then make use of the fact that $ \sqrt{a}={{a}^{\dfrac{1}{2}}} $ to proceed through the problem. We then make the necessary calculations and make use of the fact that $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ to proceed further through the problem. We then make the necessary calculations to get the required answer for the given problem.

Complete step by step answer:
According to the problem, we are asked to find the result of the given division $ \dfrac{\sqrt{64}}{\sqrt{81}} $ .
Let us assume $ d=\dfrac{\sqrt{64}}{\sqrt{81}} $ ---(1).
We know that $ \sqrt{a}={{a}^{\dfrac{1}{2}}} $ . Let us use this result in equation (1).
 $ \Rightarrow d=\dfrac{{{\left( 64 \right)}^{\dfrac{1}{2}}}}{{{\left( 81 \right)}^{\dfrac{1}{2}}}} $ ---(2).
We know that $ 64={{8}^{2}} $ and $ 81={{9}^{2}} $ . Let us use these results in equation (2).
 $ \Rightarrow d=\dfrac{{{\left( {{8}^{2}} \right)}^{\dfrac{1}{2}}}}{{{\left( {{9}^{2}} \right)}^{\dfrac{1}{2}}}} $ ---(3).
From laws of exponents, we know that $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ . Let us use this result in equation (3).
 $ \Rightarrow d=\dfrac{\left( {{8}^{2\times \dfrac{1}{2}}} \right)}{\left( {{9}^{2\times \dfrac{1}{2}}} \right)} $ .
 $ \Rightarrow d=\dfrac{\left( {{8}^{1}} \right)}{\left( {{9}^{1}} \right)} $ .
 $ \Rightarrow d=\dfrac{8}{9} $ .
So, we have found the simplified form of the given division $ \dfrac{\sqrt{64}}{\sqrt{81}} $ as $ \dfrac{8}{9} $ .
$ \therefore $ The simplified form of the given division $ \dfrac{\sqrt{64}}{\sqrt{81}} $ is $ \dfrac{8}{9} $ .

Note:
 We should perform each step carefully to avoid confusion and calculation mistakes. We can also solve this problem as shown below:
We have $ d=\dfrac{\sqrt{64}}{\sqrt{81}} $ ---(4).
We know that $ {{2}^{6}}=64 $ , $ {{3}^{4}}=81 $ . Let us use these results in equation (4).
 $ \Rightarrow d=\dfrac{\sqrt{{{2}^{6}}}}{\sqrt{{{3}^{4}}}} $ ---(5).
We know that $ \sqrt{a}={{a}^{\dfrac{1}{2}}} $ . Let us use this result in equation (5).
 $ \Rightarrow d=\dfrac{{{\left( {{2}^{6}} \right)}^{\dfrac{1}{2}}}}{{{\left( {{3}^{4}} \right)}^{\dfrac{1}{2}}}} $ ---(6).
From laws of exponents, we know that $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ . Let us use this result in equation (3).
 $ \Rightarrow d=\dfrac{\left( {{2}^{6\times \dfrac{1}{2}}} \right)}{\left( {{3}^{4\times \dfrac{1}{2}}} \right)} $ .
 $ \Rightarrow d=\dfrac{\left( {{2}^{3}} \right)}{\left( {{3}^{2}} \right)} $ ---(7).
We know that $ {{2}^{3}}=8 $ , $ {{3}^{2}}=9 $ . Let us use these results in equation (7).
 $ \Rightarrow d=\dfrac{8}{9} $ .