Question
Answer
$$\dfrac{-7}{5\sqrt{3}+9}=\dfrac{35\sqrt{3}-63}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-7}{5\sqrt{3}+9}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-7}{5\sqrt{3}+9}\frac{5\sqrt{3}-9}{5\sqrt{3}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-35\sqrt{3}+63}{75-45\sqrt{3}+45\sqrt{3}-81} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-35\sqrt{3}+63}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{35\sqrt{3}-63}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 5 \sqrt{3}-9} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ -7 } \cdot \left( 5 \sqrt{3}-9\right) = \color{blue}{-7} \cdot 5 \sqrt{3}\color{blue}{-7} \cdot-9 = \\ = - 35 \sqrt{3} + 63 $$ Simplify denominator. $$ \color{blue}{ \left( 5 \sqrt{3} + 9\right) } \cdot \left( 5 \sqrt{3}-9\right) = \color{blue}{ 5 \sqrt{3}} \cdot 5 \sqrt{3}+\color{blue}{ 5 \sqrt{3}} \cdot-9+\color{blue}{9} \cdot 5 \sqrt{3}+\color{blue}{9} \cdot-9 = \\ = 75- 45 \sqrt{3} + 45 \sqrt{3}-81 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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