If \frac{x-y}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{\sqrt{x}} \quad (x \gt 0, y \gt 0) then what is the value of \frac{x}{y}?

Explanation

Factor the expression: \frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{xy}(\sqrt{x}+\sqrt{y})} = \frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}} = \frac{1}{\sqrt{x}}. Cross multiplying gives x - \sqrt{xy} = \sqrt{y}\sqrt{x} \implies x = 2\sqrt{xy}. Squaring yields x^2 = 4xy \implies x = 4y. So \frac{x}{y} = 4.

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