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