Question
Answer
$$\dfrac{2+3\sqrt{3}}{3\sqrt{3}+2}=1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2+3\sqrt{3}}{3\sqrt{3}+2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2+3\sqrt{3}}{3\sqrt{3}+2}\frac{3\sqrt{3}-2}{3\sqrt{3}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{3}-4+27-6\sqrt{3}}{27-6\sqrt{3}+6\sqrt{3}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{23}{23} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}} \frac{ 23 : \color{orangered}{ 23 } }{ 23 : \color{orangered}{ 23 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{1}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}1\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{3}-2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 + 3 \sqrt{3}\right) } \cdot \left( 3 \sqrt{3}-2\right) = \color{blue}{2} \cdot 3 \sqrt{3}+\color{blue}{2} \cdot-2+\color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot-2 = \\ = 6 \sqrt{3}-4 + 27- 6 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{3} + 2\right) } \cdot \left( 3 \sqrt{3}-2\right) = \color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot-2+\color{blue}{2} \cdot 3 \sqrt{3}+\color{blue}{2} \cdot-2 = \\ = 27- 6 \sqrt{3} + 6 \sqrt{3}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both the top and bottom numbers by $ \color{orangered}{ 23 } $. |
| ⑤ | Remove 1 from denominator. |
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