From Förberedande kurs i matematik 1

We multiply the top and bottom of the fraction by the conjugate of the denominator, \displaystyle \sqrt{7}+\sqrt{5} , and see what it leads to:

\displaystyle \begin{align} & \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\centerdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left( 5\sqrt{7}-7\sqrt{5} \right)\centerdot \left( \sqrt{7}+\sqrt{5} \right)}{\left( \sqrt{7} \right)^{2}-\left( \sqrt{5} \right)^{2}} \\ & =\frac{5\sqrt{7}\centerdot \sqrt{7}+5\sqrt{5}\centerdot \sqrt{7}-7\sqrt{5}\centerdot \sqrt{7}-7\sqrt{5}\centerdot \sqrt{5}}{7-5} \\ & =\frac{5\left( \sqrt{7} \right)^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\left( \sqrt{5} \right)^{2}}{2} \\ & =\frac{5\centerdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\centerdot 5}{2} \\ & =\frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}=\frac{\left( 5-7 \right)\sqrt{5}\sqrt{7}}{2}=\frac{-2\sqrt{5}\sqrt{7}}{2} \\ & =-\sqrt{35} \\ \end{align}