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