Answer
$$\frac{2-\sqrt{6}}{4+\sqrt{3}}=\frac{8-2\sqrt{3}-4\sqrt{6}+3\sqrt{2}}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-\sqrt{6}}{4+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-\sqrt{6}}{4+\sqrt{3}}\frac{4-\sqrt{3}}{4-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8-2\sqrt{3}-4\sqrt{6}+3\sqrt{2}}{16-4\sqrt{3}+4\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8-2\sqrt{3}-4\sqrt{6}+3\sqrt{2}}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2- \sqrt{6}\right) } \cdot \left( 4- \sqrt{3}\right) = \color{blue}{2} \cdot4+\color{blue}{2} \cdot- \sqrt{3}\color{blue}{- \sqrt{6}} \cdot4\color{blue}{- \sqrt{6}} \cdot- \sqrt{3} = \\ = 8- 2 \sqrt{3}- 4 \sqrt{6} + 3 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 4 + \sqrt{3}\right) } \cdot \left( 4- \sqrt{3}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot4+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 16- 4 \sqrt{3} + 4 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
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