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Question Number 191529 by MATHEMATICSAM last updated on 25/Apr/23

$$\mathrm{If}\frac{x-y}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{\sqrt{x}};(x>0\mathrm{and}y>0)\mathrm{then}$$ $$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{x}{y}.$$

Answered by mehdee42 last updated on 25/Apr/23

$$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{xy}(\sqrt{x}+\sqrt{y})}=\frac{1}{\sqrt{x}}$$ $$x-\sqrt{xy}=\sqrt{xy}\Rightarrow x=2\sqrt{xy}\Rightarrow \frac{x}{y}=4✓$$ $$$$

Answered by Rasheed.Sindhi last updated on 25/Apr/23

$$\frac{x-y}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{\sqrt{x}}$$ $$\frac{{\left(\sqrt{x}\right)}^2-{\left(\sqrt{y}\right)}^2}{{\left(\sqrt{x}\right)}^2\sqrt{y}+{\left(\sqrt{y}\right)}^2\sqrt{x}}=\frac{1}{\sqrt{x}}$$ $$\frac{(\sqrt{x}-\sqrt{y})\cancel{(\sqrt{x}+\sqrt{y})}}{\sqrt{x}\sqrt{y}\cancel{(\sqrt{x}+\sqrt{y})}}=\frac{1}{\sqrt{x}}$$ $$\frac{(\sqrt{x}-\sqrt{y})}{\sqrt{y}}=1$$ $$\sqrt{x}=2\sqrt{y}$$ $$\frac{\sqrt{x}}{\sqrt{y}}=2$$ $$\frac{x}{y}=4$$

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