Question
Answer
$$\dfrac{4+\sqrt{3}}{5+\sqrt{12}}=\dfrac{14-3\sqrt{3}}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4+\sqrt{3}}{5+\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4+\sqrt{3}}{5+\sqrt{12}}\frac{5-\sqrt{12}}{5-\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{20-8\sqrt{3}+5\sqrt{3}-6}{25-10\sqrt{3}+10\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14-3\sqrt{3}}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 5- \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 4 + \sqrt{3}\right) } \cdot \left( 5- \sqrt{12}\right) = \color{blue}{4} \cdot5+\color{blue}{4} \cdot- \sqrt{12}+\color{blue}{ \sqrt{3}} \cdot5+\color{blue}{ \sqrt{3}} \cdot- \sqrt{12} = \\ = 20- 8 \sqrt{3} + 5 \sqrt{3}-6 $$ Simplify denominator. $$ \color{blue}{ \left( 5 + \sqrt{12}\right) } \cdot \left( 5- \sqrt{12}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot- \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot5+\color{blue}{ \sqrt{12}} \cdot- \sqrt{12} = \\ = 25- 10 \sqrt{3} + 10 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
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