Question
Answer
$$\dfrac{\sqrt{7}}{\sqrt{35}-\sqrt{6}}=\dfrac{7\sqrt{5}+\sqrt{42}}{29}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{7}}{\sqrt{35}-\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{7}}{\sqrt{35}-\sqrt{6}}\frac{\sqrt{35}+\sqrt{6}}{\sqrt{35}+\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7\sqrt{5}+\sqrt{42}}{35+\sqrt{210}-\sqrt{210}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7\sqrt{5}+\sqrt{42}}{29}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{35} + \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{7} } \cdot \left( \sqrt{35} + \sqrt{6}\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{35}+\color{blue}{ \sqrt{7}} \cdot \sqrt{6} = \\ = 7 \sqrt{5} + \sqrt{42} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{35}- \sqrt{6}\right) } \cdot \left( \sqrt{35} + \sqrt{6}\right) = \color{blue}{ \sqrt{35}} \cdot \sqrt{35}+\color{blue}{ \sqrt{35}} \cdot \sqrt{6}\color{blue}{- \sqrt{6}} \cdot \sqrt{35}\color{blue}{- \sqrt{6}} \cdot \sqrt{6} = \\ = 35 + \sqrt{210}- \sqrt{210}-6 $$ |
| ③ | Simplify numerator and denominator |
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