Question : Simplify the expression $\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}$, where $x=2$ and $y=3$.
Option 1: $2\sqrt{6}-6$
Option 2: $\sqrt{6}-5$
Option 3: $5-2\sqrt{6}$
Option 4: $2\sqrt{6}-5$
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Correct Answer: $2\sqrt{6}-5$
Solution :
$\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
Rationalising the denominator,
$=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}} \times \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$
$=\frac{(\sqrt{x}-\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})\times (\sqrt{x}-\sqrt{y})}$
$=\frac{(\sqrt{x}-\sqrt{y})^2}{x-y}$
Putting $x=2$ and $y=3$ in the equation.
$=\frac{(\sqrt2 -\sqrt3)^2}{2-3}$
$=\frac{2+3-2\sqrt6}{-1}$
$=-5+2\sqrt6$
Hence, the correct answer is $2\sqrt6-5$.
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