Simplify $\frac{{27}^{\frac{2}{3}}}{9^{\frac{1}{2}}\cdot 3^{\frac{3}{2}}}$

The correct option is B

1÷√3

Given expression is $\frac{{27}^{\frac{2}{3}}}{9^{\frac{1}{2}}\cdot 3^{\frac{3}{2}}}$

Since, $27$ is a perfect cube of the number $3$, it can be represented as $3\times 3\times 3$.

$\Rightarrow {\left(27\right)}^{\frac{2}{3}}={\left(3^3\right)}^{\frac{2}{3}}=3^2$.

In the denominator part, we have [${\left(9\right)}^{\frac{1}{2}}\cdot {\left(3\right)}^{\frac{3}{2}}$].

$9^{\frac{1}{2}}{3}^{\frac{3}{2}}={\left(3^2\right)}^{\frac{1}{2}}{3}^{\frac{3}{2}}=3^{\frac{5}{2}}$

So, we end up dividing $3^2$ by $3^{\frac{5}{2}}$ using the formula $a^m÷a^n$ gives us $a^{m-n}$ which gives $3^{\frac{-1}{2}}$ which is nothing but $3^0÷3^{\frac{1}{2}}$ which is $1÷3^{\frac{1}{2}}=\frac{1}{\sqrt{3}}$