Question
Answer
$$\dfrac{6}{\sqrt{27}+11}=\dfrac{-18\sqrt{3}+66}{94}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{\sqrt{27}+11}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{\sqrt{27}+11}\frac{\sqrt{27}-11}{\sqrt{27}-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18\sqrt{3}-66}{27-33\sqrt{3}+33\sqrt{3}-121} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{18\sqrt{3}-66}{-94} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-18\sqrt{3}+66}{94}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{27}-11} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( \sqrt{27}-11\right) = \color{blue}{6} \cdot \sqrt{27}+\color{blue}{6} \cdot-11 = \\ = 18 \sqrt{3}-66 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{27} + 11\right) } \cdot \left( \sqrt{27}-11\right) = \color{blue}{ \sqrt{27}} \cdot \sqrt{27}+\color{blue}{ \sqrt{27}} \cdot-11+\color{blue}{11} \cdot \sqrt{27}+\color{blue}{11} \cdot-11 = \\ = 27- 33 \sqrt{3} + 33 \sqrt{3}-121 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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