Question
Answer
$$\dfrac{2\sqrt{3}}{3\sqrt{7}-9}=-\dfrac{\sqrt{21}+3\sqrt{3}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{3}}{3\sqrt{7}-9}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{3}}{3\sqrt{7}-9}\frac{3\sqrt{7}+9}{3\sqrt{7}+9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{21}+18\sqrt{3}}{63+27\sqrt{7}-27\sqrt{7}-81} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{21}+18\sqrt{3}}{-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{21}+3\sqrt{3}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{\sqrt{21}+3\sqrt{3}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{7} + 9} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 \sqrt{3} } \cdot \left( 3 \sqrt{7} + 9\right) = \color{blue}{ 2 \sqrt{3}} \cdot 3 \sqrt{7}+\color{blue}{ 2 \sqrt{3}} \cdot9 = \\ = 6 \sqrt{21} + 18 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{7}-9\right) } \cdot \left( 3 \sqrt{7} + 9\right) = \color{blue}{ 3 \sqrt{7}} \cdot 3 \sqrt{7}+\color{blue}{ 3 \sqrt{7}} \cdot9\color{blue}{-9} \cdot 3 \sqrt{7}\color{blue}{-9} \cdot9 = \\ = 63 + 27 \sqrt{7}- 27 \sqrt{7}-81 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 6. |
| ⑤ | Place a negative sign in front of a fraction. |
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