Question
Answer
$$\dfrac{6\sqrt{3}}{\sqrt{17}-\sqrt{11}}=\sqrt{51}+\sqrt{33}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6\sqrt{3}}{\sqrt{17}-\sqrt{11}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6\sqrt{3}}{\sqrt{17}-\sqrt{11}}\frac{\sqrt{17}+\sqrt{11}}{\sqrt{17}+\sqrt{11}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{51}+6\sqrt{33}}{17+\sqrt{187}-\sqrt{187}-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{51}+6\sqrt{33}}{6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{51}+\sqrt{33}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{51}+\sqrt{33}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{17} + \sqrt{11}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 \sqrt{3} } \cdot \left( \sqrt{17} + \sqrt{11}\right) = \color{blue}{ 6 \sqrt{3}} \cdot \sqrt{17}+\color{blue}{ 6 \sqrt{3}} \cdot \sqrt{11} = \\ = 6 \sqrt{51} + 6 \sqrt{33} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{17}- \sqrt{11}\right) } \cdot \left( \sqrt{17} + \sqrt{11}\right) = \color{blue}{ \sqrt{17}} \cdot \sqrt{17}+\color{blue}{ \sqrt{17}} \cdot \sqrt{11}\color{blue}{- \sqrt{11}} \cdot \sqrt{17}\color{blue}{- \sqrt{11}} \cdot \sqrt{11} = \\ = 17 + \sqrt{187}- \sqrt{187}-11 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 6. |
| ⑤ | Remove 1 from denominator. |
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