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$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}$ is equal to :

Answer:

Explanation:

$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}})\times\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$
$=\frac{(\sqrt{7}+\sqrt{5})^{2}}{\sqrt{7}^{2}-\sqrt{5}^{2}}=\frac{(\sqrt{7}+\sqrt{5})^{2}}{7-5}$
$=\frac{7+5+2\sqrt{35}}{2}=\frac{12+2\sqrt{35}}{2}=6+\sqrt{35}$