Question
Answer
$$\dfrac{6-\sqrt{2}}{6+\sqrt{2}}=\dfrac{19-6\sqrt{2}}{17}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6-\sqrt{2}}{6+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6-\sqrt{2}}{6+\sqrt{2}}\frac{6-\sqrt{2}}{6-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{36-6\sqrt{2}-6\sqrt{2}+2}{36-6\sqrt{2}+6\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{38-12\sqrt{2}}{34} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{19-6\sqrt{2}}{17}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 6- \sqrt{2}\right) } \cdot \left( 6- \sqrt{2}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- \sqrt{2}\color{blue}{- \sqrt{2}} \cdot6\color{blue}{- \sqrt{2}} \cdot- \sqrt{2} = \\ = 36- 6 \sqrt{2}- 6 \sqrt{2} + 2 $$ Simplify denominator. $$ \color{blue}{ \left( 6 + \sqrt{2}\right) } \cdot \left( 6- \sqrt{2}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot6+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 36- 6 \sqrt{2} + 6 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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