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