+527 Compute the sum
\frac{1}{\sqrt{36} + \sqrt{42}} + \frac{1}{\sqrt{39} + \sqrt{45}}
+9673 \(\displaystyle \frac{1}{\sqrt{36} + \sqrt{42}} + \frac{1}{\sqrt{39} + \sqrt{45}}\\ =\dfrac{\sqrt{42} - \sqrt{36}}{42 - 36} + \dfrac{\sqrt{45} - \sqrt{39}}{45 - 39}\\ = \dfrac16 \left(\sqrt{45} + \sqrt{42} - \sqrt{39} - \sqrt{36}\right)\\ = \dfrac{3\sqrt 5 + \sqrt{42} - \sqrt{39} - 6}6\)
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