Question
Answer
$$\dfrac{3}{4\sqrt{6}-2\sqrt{4}}=\dfrac{12\sqrt{6}+12}{80}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3}{4\sqrt{6}-2\sqrt{4}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3}{4\sqrt{6}-2\sqrt{4}}\frac{4\sqrt{6}+2\sqrt{4}}{4\sqrt{6}+2\sqrt{4}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12\sqrt{6}+12}{96+16\sqrt{6}-16\sqrt{6}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12\sqrt{6}+12}{80}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 \sqrt{6} + 2 \sqrt{4}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 } \cdot \left( 4 \sqrt{6} + 2 \sqrt{4}\right) = \color{blue}{3} \cdot 4 \sqrt{6}+\color{blue}{3} \cdot 2 \sqrt{4} = \\ = 12 \sqrt{6} + 12 $$ Simplify denominator. $$ \color{blue}{ \left( 4 \sqrt{6}- 2 \sqrt{4}\right) } \cdot \left( 4 \sqrt{6} + 2 \sqrt{4}\right) = \color{blue}{ 4 \sqrt{6}} \cdot 4 \sqrt{6}+\color{blue}{ 4 \sqrt{6}} \cdot 2 \sqrt{4}\color{blue}{- 2 \sqrt{4}} \cdot 4 \sqrt{6}\color{blue}{- 2 \sqrt{4}} \cdot 2 \sqrt{4} = \\ = 96 + 16 \sqrt{6}- 16 \sqrt{6}-16 $$ |
| ③ | Simplify numerator and denominator |
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