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$\frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$ is equal to
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The correct answer is:
$\sqrt{7}$
$\begin{aligned} & \frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}} \\ & =\frac{[\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}]^2}{(8+\sqrt{28})-(8-\sqrt{28})} \\ & =\frac{8+\sqrt{28}+8-\sqrt{28}+2 \sqrt{64-28}}{2 \sqrt{28}} \\ & =\frac{16+2 \sqrt{36}}{2 \cdot 2 \sqrt{7}}=\frac{16+12}{4 \sqrt{7}}=\frac{28}{4 \sqrt{7}}=\sqrt{7}\end{aligned}$
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