Question
Answer
$$\dfrac{7\sqrt{3}}{5-2\sqrt{6}}=35\sqrt{3}+42\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7\sqrt{3}}{5-2\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7\sqrt{3}}{5-2\sqrt{6}}\frac{5+2\sqrt{6}}{5+2\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{35\sqrt{3}+42\sqrt{2}}{25+10\sqrt{6}-10\sqrt{6}-24} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{35\sqrt{3}+42\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}35\sqrt{3}+42\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 5 + 2 \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 \sqrt{3} } \cdot \left( 5 + 2 \sqrt{6}\right) = \color{blue}{ 7 \sqrt{3}} \cdot5+\color{blue}{ 7 \sqrt{3}} \cdot 2 \sqrt{6} = \\ = 35 \sqrt{3} + 42 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 5- 2 \sqrt{6}\right) } \cdot \left( 5 + 2 \sqrt{6}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot 2 \sqrt{6}\color{blue}{- 2 \sqrt{6}} \cdot5\color{blue}{- 2 \sqrt{6}} \cdot 2 \sqrt{6} = \\ = 25 + 10 \sqrt{6}- 10 \sqrt{6}-24 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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