Answer
$\dfrac{1}{\sqrt{32x}}=\dfrac{\sqrt{2x}}{8x}$
Work Step by Step
$\dfrac{1}{\sqrt{32x}}$
Multiply the fraction by $\dfrac{\sqrt{32x}}{\sqrt{32x}}$:
$\dfrac{1}{\sqrt{32x}}=\dfrac{1}{\sqrt{32x}}\cdot\dfrac{\sqrt{32x}}{\sqrt{32x}}=\dfrac{\sqrt{32x}}{\sqrt{(32x)^{2}}}=\dfrac{\sqrt{32x}}{32x}=...$
Rewrite the expression as $\dfrac{\sqrt{16\cdot2\cdot x}}{32x}$ and simplify:
$...=\dfrac{\sqrt{16\cdot2\cdot x}}{32x}=\dfrac{4\sqrt{2x}}{32x}=\dfrac{\sqrt{2x}}{8x}$
