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Work Shown:
[tex]\frac{2\sqrt{72}}{\sqrt{8}+\sqrt{2}}\\\\\frac{2\sqrt{36*2}}{\sqrt{4*2}+\sqrt{2}}\\\\\frac{2\sqrt{36}*\sqrt{2}}{\sqrt{4}*\sqrt{2}+\sqrt{2}}\\\\\frac{2*6*\sqrt{2}}{2*\sqrt{2}+\sqrt{2}}\\\\\frac{12\sqrt{2}}{2\sqrt{2}+\sqrt{2}}\\\\\frac{12\sqrt{2}}{3\sqrt{2}}\\\\\frac{12}{3}\\\\4[/tex]
Note in step 2, I factored each number in the square root to pull out the largest perfect square factor. From there, I used the rule that [tex]\sqrt{A*B} = \sqrt{A}*\sqrt{B}[/tex] to break up the roots.
Answer:
4
Step-by-step explanation:
We first simplify $\sqrt{72}$ and $\sqrt{8}$:\begin{align*}
\sqrt{72} &=\sqrt{36\cdot 2} = \sqrt{36}\cdot \sqrt{2} = 6\sqrt{2},\\
\sqrt{8} &= \sqrt{4\cdot 2} = \sqrt{4}\cdot \sqrt{2} = 2\sqrt{2}.
\end{align*}Then, we have
\[\dfrac{2\sqrt{72}}{\sqrt{8}+\sqrt{2}} = \dfrac{2\left(6\sqrt{2}\right)}{2\sqrt{2} + \sqrt{2}}
= \dfrac{12\sqrt{2}}{3\sqrt{2}} = \dfrac{12}{3}\cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \boxed{4}.\]
