Question
Answer
$$\dfrac{10}{\sqrt{59}-\sqrt{41}}=\dfrac{10\sqrt{59}+10\sqrt{41}}{18}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{10}{\sqrt{59}-\sqrt{41}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{10}{\sqrt{59}-\sqrt{41}}\frac{\sqrt{59}+\sqrt{41}}{\sqrt{59}+\sqrt{41}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{10\sqrt{59}+10\sqrt{41}}{59+\sqrt{2419}-\sqrt{2419}-41} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{10\sqrt{59}+10\sqrt{41}}{18}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{59} + \sqrt{41}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 10 } \cdot \left( \sqrt{59} + \sqrt{41}\right) = \color{blue}{10} \cdot \sqrt{59}+\color{blue}{10} \cdot \sqrt{41} = \\ = 10 \sqrt{59} + 10 \sqrt{41} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{59}- \sqrt{41}\right) } \cdot \left( \sqrt{59} + \sqrt{41}\right) = \color{blue}{ \sqrt{59}} \cdot \sqrt{59}+\color{blue}{ \sqrt{59}} \cdot \sqrt{41}\color{blue}{- \sqrt{41}} \cdot \sqrt{59}\color{blue}{- \sqrt{41}} \cdot \sqrt{41} = \\ = 59 + \sqrt{2419}- \sqrt{2419}-41 $$ |
| ③ | Simplify numerator and denominator |
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