Answer
$$\frac{(\sqrt{18}+\sqrt{2})^2}{\sqrt{8}-2}=\frac{256+32\sqrt{2}}{62}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(\sqrt{18}+\sqrt{2})^2}{\sqrt{8}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{18+6+6+2}{8-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{32}{8-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{32}{8-\sqrt{2}}\frac{8+\sqrt{2}}{8+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{256+32\sqrt{2}}{64+8\sqrt{2}-8\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{256+32\sqrt{2}}{62}\end{aligned} $$ | |
| ① | $$ (\sqrt{18}+\sqrt{2})^2 = \left( \sqrt{18} + \sqrt{2} \right) \cdot \left( \sqrt{18} + \sqrt{2} \right) = 18 + 6 + 6 + 2 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 8 + \sqrt{2}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 32 } \cdot \left( 8 + \sqrt{2}\right) = \color{blue}{32} \cdot8+\color{blue}{32} \cdot \sqrt{2} = \\ = 256 + 32 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 8- \sqrt{2}\right) } \cdot \left( 8 + \sqrt{2}\right) = \color{blue}{8} \cdot8+\color{blue}{8} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot8\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 64 + 8 \sqrt{2}- 8 \sqrt{2}-2 $$ |
| ⑤ | Simplify numerator and denominator |
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