Find the value of $$A=\dfrac{3\sqrt{8+2\sqrt7}}{\sqrt{8-2\sqrt{7}}}-\dfrac{\sqrt{2\left(3+\sqrt7\right)}}{\sqrt{3-\sqrt7}}.$$
Answer: $A=1$
Since $$8+2\sqrt7=8+2\cdot1\cdot\sqrt{7}=1^2+2\cdot1\cdot\sqrt7+\left(\sqrt{7}\right)^2=(1+\sqrt{7})^2,\\8-2\sqrt{7}=(1-\sqrt{7})^2,$$ then $$A=\dfrac{3(1+\sqrt{7})}{\sqrt{7}-1}-\dfrac{\sqrt{2\left(3+\sqrt7\right)}}{\sqrt{3-\sqrt7}}.$$ What can we do with the second fraction?
$\begin{align}&A = \frac{3\sqrt{8+2\sqrt7}}{\sqrt{8-2\sqrt7}}\frac{\sqrt{8+2\sqrt7}}{\sqrt{8+2\sqrt7}}- \frac{\sqrt2\sqrt{3+\sqrt7}}{\sqrt{3-\sqrt7}}\frac{\sqrt{3+\sqrt7}}{\sqrt{3+\sqrt7}}\\\Rightarrow&A=\frac{3(8+2\sqrt7)}{6}-\sqrt2\left(\frac{3 +\sqrt7}{\sqrt2}\right) = 4+\sqrt7-3-\sqrt7=1\end{align}$