Question
Simplify:$\frac{2}{\sqrt5+\sqrt3}+\frac{1}{\sqrt3+\sqrt2}-\frac{3}{\sqrt5+\sqrt2}$
✓
Answer
We know that rationalization factor for $\sqrt5+\sqrt3,\ \sqrt3+\sqrt2,$ and $\sqrt5+\sqrt2$ are $\sqrt5+\sqrt3,\ \sqrt3-\sqrt2$ and $\sqrt5-\sqrt2$ respectively. We will multiply numerator and denominator of the given expression $\frac{2}{\sqrt5+\sqrt3},\ \frac{1}{\sqrt3+\sqrt2}$ and $\frac{3}{\sqrt5+\sqrt2}$ by $2-\sqrt3,$ $\sqrt5+\sqrt3$ and $2+\sqrt5$ respectively, to get$\frac{2}{\sqrt5+\sqrt2}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}+\frac{1}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}-\frac{3}{\sqrt5+\sqrt2}\times\frac{\sqrt5-\sqrt2}{\sqrt5-\sqrt2}\\ \ \ =\frac{2\sqrt5-2\sqrt3}{5-3}+\frac{\sqrt3-\sqrt2}{3-2}-\frac{3\sqrt5-3\sqrt2}{5-2}$
$=\frac{2\sqrt5-2\sqrt{3}}{2}+\frac{\sqrt{3}-\sqrt2}{1}-\frac{3\sqrt5-3\sqrt2}{3}$
$=\sqrt5-\sqrt3+\sqrt3-\sqrt2-\sqrt5+\sqrt2$
$=0$
Hence the given expression is simplified to 0.
$=\frac{2\sqrt5-2\sqrt{3}}{2}+\frac{\sqrt{3}-\sqrt2}{1}-\frac{3\sqrt5-3\sqrt2}{3}$
$=\sqrt5-\sqrt3+\sqrt3-\sqrt2-\sqrt5+\sqrt2$
$=0$
Hence the given expression is simplified to 0.
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