We know that rationalization factor for $3+\sqrt5$ and $3-\sqrt5$ are $3-\sqrt5$ and $3+\sqrt5$ respectively. We will multiply numerator and denominator of the given expression $\frac{7+3\sqrt5}{3+\sqrt5}$ and $\frac{7-3\sqrt5}{3+\sqrt5}$ by $3-\sqrt5$ and $3+\sqrt5$ respectively. to get$\frac{7+3\sqrt5}{3+\sqrt5}\times\frac{3-\sqrt5}{3-\sqrt5}-\frac{7-3\sqrt5}{3-\sqrt5}\times\frac{3+\sqrt5}{3+\sqrt5}\\ \ =\frac{7\times3-7\times\sqrt5+9\times\sqrt5-3\times\big(\sqrt5\big)^2}{(3)^2-\big(\sqrt5\big)^2}-\frac{7\times3+7\times\sqrt5-9\times\sqrt5-3\times\big(\sqrt5\big)^2}{(3)^2-\big(\sqrt5\big)^2}$
$=\frac{21-7\sqrt5+9\sqrt5-3\times5}{9-5}-\frac{21+7\sqrt5-9\sqrt5-3\times5}{9-5}$
$=\frac{21+2\sqrt5-15}{4}-\frac{21-2\sqrt5-15}{4}$
$=\frac{6+2\sqrt5-6+2\sqrt5}{4}$
$=\frac{4\sqrt5}{4}$
$=\sqrt5$
Hence the given expression is simplified to $\sqrt5.$
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