Question
Answer
$$\dfrac{9}{\sqrt{3}+\sqrt{12}}=\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{9}{\sqrt{3}+\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9}{\sqrt{3}+\sqrt{12}}\frac{\sqrt{3}-\sqrt{12}}{\sqrt{3}-\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{9\sqrt{3}-18\sqrt{3}}{3-6+6-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-9\sqrt{3}}{-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\sqrt{3}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}- \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 9 } \cdot \left( \sqrt{3}- \sqrt{12}\right) = \color{blue}{9} \cdot \sqrt{3}+\color{blue}{9} \cdot- \sqrt{12} = \\ = 9 \sqrt{3}- 18 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + \sqrt{12}\right) } \cdot \left( \sqrt{3}- \sqrt{12}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot \sqrt{3}+\color{blue}{ \sqrt{12}} \cdot- \sqrt{12} = \\ = 3-6 + 6-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 9. |
| ⑤ | Multiply both numerator and denominator by -1. |
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