Question
Simplify: $\frac{1}{2+\sqrt3}+\frac{2}{\sqrt5-\sqrt3}+\frac{1}{2-\sqrt5}$
✓
Answer
We know that rationalization factor for $2+\sqrt3,\ \sqrt5-\sqrt3,$ and $2-\sqrt5$ are $2-\sqrt3,\ \sqrt5+\sqrt3$ and $2+\sqrt5$ respectively.
We will multiply numerator and denominator of the given expression $\frac{1}{2+\sqrt3},\ \frac{2}{\sqrt5-\sqrt3}$ and $\frac{1}{2-\sqrt5}$ by $2-\sqrt3.$
$\sqrt5+\sqrt3$ and $2+\sqrt5$ respectively,
to get $\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}+\frac{2}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}+\frac{1}{2-\sqrt5}\times\frac{2+\sqrt5}{2+\sqrt5}$
$\ \ =\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}+\frac{2\sqrt5+2\sqrt3}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}+\frac{2-\sqrt5}{(2)^2-\big(\sqrt5\big)^2}$
$=\frac{2-\sqrt{3}}{1}+\frac{2\sqrt{5}+2\sqrt3}{5-3}+\frac{2+\sqrt5}{4-5}$
$=\frac{2-\sqrt{3}}{1}+\frac{2\sqrt{5}+2\sqrt3}{2}+\frac{2+\sqrt5}{-1}$
$=2-\sqrt3+\sqrt5+\sqrt3-\sqrt5-2$
$=0$
Hence the given expression is simplified to $0$.
We will multiply numerator and denominator of the given expression $\frac{1}{2+\sqrt3},\ \frac{2}{\sqrt5-\sqrt3}$ and $\frac{1}{2-\sqrt5}$ by $2-\sqrt3.$
$\sqrt5+\sqrt3$ and $2+\sqrt5$ respectively,
to get $\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}+\frac{2}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}+\frac{1}{2-\sqrt5}\times\frac{2+\sqrt5}{2+\sqrt5}$
$\ \ =\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}+\frac{2\sqrt5+2\sqrt3}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}+\frac{2-\sqrt5}{(2)^2-\big(\sqrt5\big)^2}$
$=\frac{2-\sqrt{3}}{1}+\frac{2\sqrt{5}+2\sqrt3}{5-3}+\frac{2+\sqrt5}{4-5}$
$=\frac{2-\sqrt{3}}{1}+\frac{2\sqrt{5}+2\sqrt3}{2}+\frac{2+\sqrt5}{-1}$
$=2-\sqrt3+\sqrt5+\sqrt3-\sqrt5-2$
$=0$
Hence the given expression is simplified to $0$.
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