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\frac{\sqrt{3}-5}{\left(\sqrt{3}+5\right)\left(\sqrt{3}-5\right)}-\frac{1}{\sqrt{3}-5}
Rationalize the denominator of \frac{1}{\sqrt{3}+5} by multiplying numerator and denominator by \sqrt{3}-5.
\frac{\sqrt{3}-5}{\left(\sqrt{3}\right)^{2}-5^{2}}-\frac{1}{\sqrt{3}-5}
Consider \left(\sqrt{3}+5\right)\left(\sqrt{3}-5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\sqrt{3}-5}{3-25}-\frac{1}{\sqrt{3}-5}
Square \sqrt{3}. Square 5.
\frac{\sqrt{3}-5}{-22}-\frac{1}{\sqrt{3}-5}
Subtract 25 from 3 to get -22.
\frac{-\sqrt{3}+5}{22}-\frac{1}{\sqrt{3}-5}
Multiply both numerator and denominator by -1.
\frac{-\sqrt{3}+5}{22}-\frac{\sqrt{3}+5}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}
Rationalize the denominator of \frac{1}{\sqrt{3}-5} by multiplying numerator and denominator by \sqrt{3}+5.
\frac{-\sqrt{3}+5}{22}-\frac{\sqrt{3}+5}{\left(\sqrt{3}\right)^{2}-5^{2}}
Consider \left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{-\sqrt{3}+5}{22}-\frac{\sqrt{3}+5}{3-25}
Square \sqrt{3}. Square 5.
\frac{-\sqrt{3}+5}{22}-\frac{\sqrt{3}+5}{-22}
Subtract 25 from 3 to get -22.
\frac{-\sqrt{3}+5}{22}-\frac{-\sqrt{3}-5}{22}
Multiply both numerator and denominator by -1.
\frac{-\sqrt{3}+5-\left(-\sqrt{3}-5\right)}{22}
Since \frac{-\sqrt{3}+5}{22} and \frac{-\sqrt{3}-5}{22} have the same denominator, subtract them by subtracting their numerators.
\frac{-\sqrt{3}+5+\sqrt{3}+5}{22}
Do the multiplications in -\sqrt{3}+5-\left(-\sqrt{3}-5\right).
\frac{10}{22}
Do the calculations in -\sqrt{3}+5+\sqrt{3}+5.
\frac{5}{11}
Reduce the fraction \frac{10}{22} to lowest terms by extracting and canceling out 2.