Question
Answer
$$\dfrac{\sqrt{21}}{5-\sqrt{4}}=\dfrac{\sqrt{21}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{21}}{5-\sqrt{4}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{21}}{5-\sqrt{4}}\frac{5+\sqrt{4}}{5+\sqrt{4}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5\sqrt{21}+2\sqrt{21}}{25+10-10-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7\sqrt{21}}{21} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{21}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 5 + \sqrt{4}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{21} } \cdot \left( 5 + \sqrt{4}\right) = \color{blue}{ \sqrt{21}} \cdot5+\color{blue}{ \sqrt{21}} \cdot \sqrt{4} = \\ = 5 \sqrt{21} + 2 \sqrt{21} $$ Simplify denominator. $$ \color{blue}{ \left( 5- \sqrt{4}\right) } \cdot \left( 5 + \sqrt{4}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot \sqrt{4}\color{blue}{- \sqrt{4}} \cdot5\color{blue}{- \sqrt{4}} \cdot \sqrt{4} = \\ = 25 + 10-10-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 7. |
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