From Förberedande kurs i matematik 1

First, we multiply the tops and bottoms of the two terms by the conjugate of their respective denominators, so that there are no root signs left in the denominators,

\displaystyle \begin{align} & \frac{1}{\sqrt{6}-\sqrt{5}}=\frac{1}{\sqrt{6}-\sqrt{5}}\centerdot \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{\sqrt{6}+\sqrt{5}}{\left( \sqrt{6} \right)^{2}-\left( \sqrt{5} \right)^{2}}=\frac{\sqrt{6}+\sqrt{5}}{6-5}=\sqrt{6}+\sqrt{5}, \\ & \frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\centerdot \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{\left( \sqrt{7} \right)^{2}-\left( \sqrt{6} \right)^{2}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6}, \\ & \\ \end{align}

Now, we can subtract the terms and simplify the result,

\displaystyle \begin{align} & \frac{1}{\sqrt{6}-\sqrt{5}}-\frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{6}+\sqrt{5}-\left( \sqrt{7}+\sqrt{6} \right) \\ & =\sqrt{6}+\sqrt{5}-\sqrt{7}-\sqrt{6}=\sqrt{5}-\sqrt{7}. \\ \end{align}