Question Number 191529 by MATHEMATICSAM last updated on 25/Apr/23
$$\mathrm{If}\:\frac{{x}\:−\:{y}}{{x}\sqrt{{y}}\:+\:{y}\sqrt{{x}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{x}}}\:;\:\left({x}\:>\:\mathrm{0}\:\mathrm{and}\:{y}\:>\:\mathrm{0}\right)\:\mathrm{then} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{{x}}{{y}}\:. \\ $$
Answered by mehdee42 last updated on 25/Apr/23
$$\frac{\left(\sqrt{{x}}−\sqrt{{y}}\right)\left(\sqrt{{x}}+\sqrt{{y}}\right)}{\:\sqrt{{xy}}\left(\sqrt{{x}}+\sqrt{{y}}\right)}=\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$${x}−\sqrt{{xy}}=\sqrt{{xy}}\Rightarrow{x}=\mathrm{2}\sqrt{{xy}}\Rightarrow\frac{{x}}{{y}}=\mathrm{4}\:\checkmark \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 25/Apr/23
$$\:\frac{{x}\:−\:{y}}{{x}\sqrt{{y}}\:+\:{y}\sqrt{{x}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$$\:\frac{\left(\sqrt{{x}}\:\right)^{\mathrm{2}} \:−\:\left(\sqrt{{y}}\:\right)^{\mathrm{2}} }{\:\left(\sqrt{{x}}\:\right)^{\mathrm{2}} \sqrt{{y}}\:+\:\left(\sqrt{{y}}\:\right)^{\mathrm{2}} \sqrt{{x}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$$\:\frac{\left(\sqrt{{x}}\:\:−\:\sqrt{{y}}\:\right)\cancel{\left(\sqrt{{x}}\:\:+\:\sqrt{{y}}\:\right)}}{\:\sqrt{{x}}\:\sqrt{{y}}\:\cancel{\left(\sqrt{{x}}\:+\sqrt{{y}}\:\right)}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$$\:\frac{\left(\sqrt{{x}}\:\:−\:\sqrt{{y}}\:\right)}{\:\:\sqrt{{y}}\:}\:=\:\mathrm{1} \\ $$$$\:\:\sqrt{{x}}\:=\mathrm{2}\sqrt{{y}} \\ $$$$\frac{\sqrt{{x}}\:}{\:\sqrt{{y}}}=\mathrm{2} \\ $$$$\frac{{x}}{{y}}=\mathrm{4} \\ $$