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\frac{\sqrt{6}+\sqrt{5}}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}
Rationalize the denominator of \frac{1}{\sqrt{6}-\sqrt{5}} by multiplying numerator and denominator by \sqrt{6}+\sqrt{5}.
\frac{\sqrt{6}+\sqrt{5}}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{5}\right)^{2}}
Consider \left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{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{6}+\sqrt{5}}{6-5}
Square \sqrt{6}. Square \sqrt{5}.
\frac{\sqrt{6}+\sqrt{5}}{1}
Subtract 5 from 6 to get 1.
\sqrt{6}+\sqrt{5}
Anything divided by one gives itself.