Question
Answer
$$\dfrac{6}{2-\sqrt{2}}-4\sqrt{2}=6-\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{2-\sqrt{2}}-4\sqrt{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{12+6\sqrt{2}}{4+2\sqrt{2}-2\sqrt{2}-2}-4\sqrt{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12+6\sqrt{2}}{2}-4\sqrt{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12+6\sqrt{2}-8\sqrt{2}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{12-2\sqrt{2}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{6-\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}6-\sqrt{2}\end{aligned} $$ | |
| ① | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( 2 + \sqrt{2}\right) = \color{blue}{6} \cdot2+\color{blue}{6} \cdot \sqrt{2} = \\ = 12 + 6 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{2}\right) } \cdot \left( 2 + \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot2\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 4 + 2 \sqrt{2}- 2 \sqrt{2}-2 $$ |
| ② | Simplify numerator and denominator |
| ③ | $$ \frac{12+6\sqrt{2}}{2}-4\sqrt{2}
= \frac{12+6\sqrt{2}}{2} \cdot \color{blue}{\frac{ 1 }{ 1}} - 4\sqrt{2} \cdot \color{blue}{\frac{ 2 }{ 2}}
= \frac{12+6\sqrt{2}-8\sqrt{2}}{2} $$ |
| ④ | Simplify numerator and denominator |
| ⑤ | Divide both numerator and denominator by 2. |
| ⑥ | Remove 1 from denominator. |
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