\[\frac{6^2(k-l)^2}{(k+l)^2} = \frac{36(k-l)^2}{(k+l)^2}\]
При \(k = -\sqrt{5}\) и \(l = \sqrt{7}\):
\[\frac{36(-\sqrt{5} - \sqrt{7})^2}{(-\sqrt{5} + \sqrt{7})^2} = \frac{36(\sqrt{5} + \sqrt{7})^2}{(\sqrt{7} - \sqrt{5})^2}\]
\[= 36 \cdot \frac{(\sqrt{5} + \sqrt{7})^2}{(\sqrt{7} - \sqrt{5})^2} = 36 \cdot \frac{(5 + 2\sqrt{35} + 7)}{(7 - 2\sqrt{35} + 5)} = 36 \cdot \frac{12 + 2\sqrt{35}}{12 - 2\sqrt{35}}\]
\[= 36 \cdot \frac{6 + \sqrt{35}}{6 - \sqrt{35}} = 36 \cdot \frac{(6 + \sqrt{35})(6 + \sqrt{35})}{(6 - \sqrt{35})(6 + \sqrt{35})} = 36 \cdot \frac{36 + 12\sqrt{35} + 35}{36 - 35}\]
\[= 36(71 + 12\sqrt{35}) = 2556 + 432\sqrt{35}\]
Ответ: \(2556 + 432\sqrt{35}\)