Hey,

This question is pretty straight forward.

We have

\(\frac{x}{y}\sqrt{\frac{y}{x^2}}\)

rewritten as

\(\sqrt{\frac{x}{y}} \times \sqrt{\frac{x}{y}} \times \sqrt{\frac{y}{x}}\times\sqrt{\frac{1}{x}}\)

cancelling out \(\sqrt{\frac{x}{y}}\) and \(\sqrt{\frac{y}{x}}\)

As x > 0 and y > 0 \(\sqrt{x}\) and \(\sqrt{y}\) are defined and \(\frac{x}{y}\) and \(\frac{y}{x}\) are also defined.

\(\sqrt{\frac{x}{y}} \times \sqrt{\frac{1}{x}}\)

cancelling out \(\sqrt{x}\)

\(\sqrt{\frac{1}{y}}\)

Hence E is the correct answer.
_________________

Sandy

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