Question
Answer
$$\dfrac{\dfrac{\sqrt{3}}{3}+1}{1-\dfrac{\sqrt{3}}{3}}=\sqrt{3}+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\frac{\sqrt{3}+3}{3}}{\frac{3-\sqrt{3}}{3}} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{\sqrt{3}+3}{3}\cdot\frac{3}{3-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3\sqrt{3}+9}{9-3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3\sqrt{3}+9}{9-3\sqrt{3}}\frac{9+3\sqrt{3}}{9+3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{27\sqrt{3}+27+81+27\sqrt{3}}{81+27\sqrt{3}-27\sqrt{3}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{54\sqrt{3}+108}{54} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{\sqrt{3}+2}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\sqrt{3}+2\end{aligned} $$ | |
| ① | $$ \frac{\sqrt{3}}{3}+1
= \frac{\sqrt{3}}{3} \cdot \color{blue}{\frac{ 1 }{ 1}} + 1 \cdot \color{blue}{\frac{ 3 }{ 3}}
= \frac{\sqrt{3}+3}{3} $$ |
| ② | $$ 1-\frac{\sqrt{3}}{3}
= 1 \cdot \color{blue}{\frac{ 3 }{ 3}} - \frac{\sqrt{3}}{3} \cdot \color{blue}{\frac{ 1 }{ 1}}
= \frac{3-\sqrt{3}}{3} $$ |
| ③ | $$ \color{blue}{ \left( \sqrt{3} + 3\right) } \cdot 3 = \color{blue}{ \sqrt{3}} \cdot3+\color{blue}{3} \cdot3 = \\ = 3 \sqrt{3} + 9 $$$$ \color{blue}{ 3 } \cdot \left( 3- \sqrt{3}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{3} = \\ = 9- 3 \sqrt{3} $$ |
| ④ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 9 + 3 \sqrt{3}} $$. |
| ⑤ | Multiply in a numerator. $$ \color{blue}{ \left( 3 \sqrt{3} + 9\right) } \cdot \left( 9 + 3 \sqrt{3}\right) = \color{blue}{ 3 \sqrt{3}} \cdot9+\color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{9} \cdot9+\color{blue}{9} \cdot 3 \sqrt{3} = \\ = 27 \sqrt{3} + 27 + 81 + 27 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 9- 3 \sqrt{3}\right) } \cdot \left( 9 + 3 \sqrt{3}\right) = \color{blue}{9} \cdot9+\color{blue}{9} \cdot 3 \sqrt{3}\color{blue}{- 3 \sqrt{3}} \cdot9\color{blue}{- 3 \sqrt{3}} \cdot 3 \sqrt{3} = \\ = 81 + 27 \sqrt{3}- 27 \sqrt{3}-27 $$ |
| ⑥ | Simplify numerator and denominator |
| ⑦ | Divide both numerator and denominator by 54. |
| ⑧ | Remove 1 from denominator. |
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