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\frac{5\left(\sqrt{7}-1\right)}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}-\frac{5}{\sqrt{7}-1}
Rationalize the denominator of \frac{5}{\sqrt{7}+1} by multiplying numerator and denominator by \sqrt{7}-1.
\frac{5\left(\sqrt{7}-1\right)}{\left(\sqrt{7}\right)^{2}-1^{2}}-\frac{5}{\sqrt{7}-1}
Consider \left(\sqrt{7}+1\right)\left(\sqrt{7}-1\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{5\left(\sqrt{7}-1\right)}{7-1}-\frac{5}{\sqrt{7}-1}
Square \sqrt{7}. Square 1.
\frac{5\left(\sqrt{7}-1\right)}{6}-\frac{5}{\sqrt{7}-1}
Subtract 1 from 7 to get 6.
\frac{5\left(\sqrt{7}-1\right)}{6}-\frac{5\left(\sqrt{7}+1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}
Rationalize the denominator of \frac{5}{\sqrt{7}-1} by multiplying numerator and denominator by \sqrt{7}+1.
\frac{5\left(\sqrt{7}-1\right)}{6}-\frac{5\left(\sqrt{7}+1\right)}{\left(\sqrt{7}\right)^{2}-1^{2}}
Consider \left(\sqrt{7}-1\right)\left(\sqrt{7}+1\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{5\left(\sqrt{7}-1\right)}{6}-\frac{5\left(\sqrt{7}+1\right)}{7-1}
Square \sqrt{7}. Square 1.
\frac{5\left(\sqrt{7}-1\right)}{6}-\frac{5\left(\sqrt{7}+1\right)}{6}
Subtract 1 from 7 to get 6.
\frac{5\left(\sqrt{7}-1\right)-5\left(\sqrt{7}+1\right)}{6}
Since \frac{5\left(\sqrt{7}-1\right)}{6} and \frac{5\left(\sqrt{7}+1\right)}{6} have the same denominator, subtract them by subtracting their numerators.
\frac{5\sqrt{7}-5-5\sqrt{7}-5}{6}
Do the multiplications in 5\left(\sqrt{7}-1\right)-5\left(\sqrt{7}+1\right).
\frac{-10}{6}
Do the calculations in 5\sqrt{7}-5-5\sqrt{7}-5.
-\frac{5}{3}
Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.