Question
Answer
$$\dfrac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}=\dfrac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}\frac{\sqrt{42}}{\sqrt{42}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{42}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{7} + \sqrt{6} + \sqrt{13}\right) } \cdot \sqrt{42} = \color{blue}{ \sqrt{7}} \cdot \sqrt{42}+\color{blue}{ \sqrt{6}} \cdot \sqrt{42}+\color{blue}{ \sqrt{13}} \cdot \sqrt{42} = \\ = 7 \sqrt{6} + 6 \sqrt{7} + \sqrt{546} $$ Simplify denominator. $$ \color{blue}{ 2 \sqrt{42} } \cdot \sqrt{42} = 84 $$ |
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