Question
Answer
$$\dfrac{7}{9\sqrt{7}+2\sqrt{3}}=\dfrac{63\sqrt{7}-14\sqrt{3}}{555}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{9\sqrt{7}+2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{9\sqrt{7}+2\sqrt{3}}\frac{9\sqrt{7}-2\sqrt{3}}{9\sqrt{7}-2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{63\sqrt{7}-14\sqrt{3}}{567-18\sqrt{21}+18\sqrt{21}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{63\sqrt{7}-14\sqrt{3}}{555}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 9 \sqrt{7}- 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( 9 \sqrt{7}- 2 \sqrt{3}\right) = \color{blue}{7} \cdot 9 \sqrt{7}+\color{blue}{7} \cdot- 2 \sqrt{3} = \\ = 63 \sqrt{7}- 14 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 9 \sqrt{7} + 2 \sqrt{3}\right) } \cdot \left( 9 \sqrt{7}- 2 \sqrt{3}\right) = \color{blue}{ 9 \sqrt{7}} \cdot 9 \sqrt{7}+\color{blue}{ 9 \sqrt{7}} \cdot- 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot 9 \sqrt{7}+\color{blue}{ 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 567- 18 \sqrt{21} + 18 \sqrt{21}-12 $$ |
| ③ | Simplify numerator and denominator |
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