Question
Answer
$$\dfrac{7}{6-3\sqrt{2}}=\dfrac{42+21\sqrt{2}}{18}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{6-3\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{6-3\sqrt{2}}\frac{6+3\sqrt{2}}{6+3\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{42+21\sqrt{2}}{36+18\sqrt{2}-18\sqrt{2}-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{42+21\sqrt{2}}{18}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6 + 3 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( 6 + 3 \sqrt{2}\right) = \color{blue}{7} \cdot6+\color{blue}{7} \cdot 3 \sqrt{2} = \\ = 42 + 21 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 6- 3 \sqrt{2}\right) } \cdot \left( 6 + 3 \sqrt{2}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot 3 \sqrt{2}\color{blue}{- 3 \sqrt{2}} \cdot6\color{blue}{- 3 \sqrt{2}} \cdot 3 \sqrt{2} = \\ = 36 + 18 \sqrt{2}- 18 \sqrt{2}-18 $$ |
| ③ | Simplify numerator and denominator |
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