Question
Answer
$$\dfrac{1}{2\sqrt{6}+3}=\dfrac{2\sqrt{6}-3}{15}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{2\sqrt{6}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{2\sqrt{6}+3}\frac{2\sqrt{6}-3}{2\sqrt{6}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{6}-3}{24-6\sqrt{6}+6\sqrt{6}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{6}-3}{15}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{6}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 2 \sqrt{6}-3\right) = \color{blue}{1} \cdot 2 \sqrt{6}+\color{blue}{1} \cdot-3 = \\ = 2 \sqrt{6}-3 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{6} + 3\right) } \cdot \left( 2 \sqrt{6}-3\right) = \color{blue}{ 2 \sqrt{6}} \cdot 2 \sqrt{6}+\color{blue}{ 2 \sqrt{6}} \cdot-3+\color{blue}{3} \cdot 2 \sqrt{6}+\color{blue}{3} \cdot-3 = \\ = 24- 6 \sqrt{6} + 6 \sqrt{6}-9 $$ |
| ③ | Simplify numerator and denominator |
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