Question
Answer
$$\dfrac{7+2\sqrt{5}}{6+3\sqrt{3}}=\dfrac{14-7\sqrt{3}+4\sqrt{5}-2\sqrt{15}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7+2\sqrt{5}}{6+3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7+2\sqrt{5}}{6+3\sqrt{3}}\frac{6-3\sqrt{3}}{6-3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{42-21\sqrt{3}+12\sqrt{5}-6\sqrt{15}}{36-18\sqrt{3}+18\sqrt{3}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{42-21\sqrt{3}+12\sqrt{5}-6\sqrt{15}}{9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{14-7\sqrt{3}+4\sqrt{5}-2\sqrt{15}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7 + 2 \sqrt{5}\right) } \cdot \left( 6- 3 \sqrt{3}\right) = \color{blue}{7} \cdot6+\color{blue}{7} \cdot- 3 \sqrt{3}+\color{blue}{ 2 \sqrt{5}} \cdot6+\color{blue}{ 2 \sqrt{5}} \cdot- 3 \sqrt{3} = \\ = 42- 21 \sqrt{3} + 12 \sqrt{5}- 6 \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( 6 + 3 \sqrt{3}\right) } \cdot \left( 6- 3 \sqrt{3}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot6+\color{blue}{ 3 \sqrt{3}} \cdot- 3 \sqrt{3} = \\ = 36- 18 \sqrt{3} + 18 \sqrt{3}-27 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
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