Answer
$$\frac{17}{2+\sqrt{2}-\sqrt{2}\cdot\sqrt{2}}=\frac{17\sqrt{2}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{17}{2+\sqrt{2}-\sqrt{2}\cdot\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{17}{2+\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{17}{\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}} \frac{ 17 }{\sqrt{ 2 }} \times \frac{ \color{orangered}{\sqrt{ 2 }} }{ \color{orangered}{\sqrt{ 2 }}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{17\sqrt{2}}{2}\end{aligned} $$ | |
| ① | $$ - \sqrt{4} = -1 \cdot 2 = -2 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply both top and bottom by $ \color{orangered}{ \sqrt{ 2 }}$. |
| ④ | In denominator we have $ \sqrt{ 2 } \cdot \sqrt{ 2 } = 2 $. |
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