, 2022

Evaluating, $\text{3 2}$ - 2^3}{2^3 -
3^2} = $\text{9}$ - 8}{8-9} = $\dfrac{1}{-1}$ = -1$.

Alternatively, since $2^3 - 3^2 = -(3^2 -
2^3)$, then$$\text{3 2}$ - 2^3}{2^3 -
3^2} = -1$.
Evaluating, $\$\sqrt{\phantom{\left(\right.}\hspace{-2mm}\sqrt{81}+\sqrt{9}-\sqrt{64}} = \sqrt{9}$ + 3 - 8} = $\sqrt{4}$ = 2$.
Since $\$\dfrac{1}{\sqrt{x^2}}$ + 7}} =
$\dfrac{1}{4}$$, then$$\sqrt{x^2}$ + 7} =
4$.

This means that $x^2 + 7 = 4^2 = 16$
and so $x^2 = 9$.

Since $x^2 = 9$, then $x = \pm 3$.

We can check by substitution that both of these values are
solutions.