Question
Answer
$$\dfrac{\sqrt{21}}{2+\sqrt{21}}=\dfrac{-2\sqrt{21}+21}{17}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{21}}{2+\sqrt{21}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{21}}{2+\sqrt{21}}\frac{2-\sqrt{21}}{2-\sqrt{21}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{21}-21}{4-2\sqrt{21}+2\sqrt{21}-21} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{21}-21}{-17} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2\sqrt{21}+21}{17}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- \sqrt{21}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{21} } \cdot \left( 2- \sqrt{21}\right) = \color{blue}{ \sqrt{21}} \cdot2+\color{blue}{ \sqrt{21}} \cdot- \sqrt{21} = \\ = 2 \sqrt{21}-21 $$ Simplify denominator. $$ \color{blue}{ \left( 2 + \sqrt{21}\right) } \cdot \left( 2- \sqrt{21}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- \sqrt{21}+\color{blue}{ \sqrt{21}} \cdot2+\color{blue}{ \sqrt{21}} \cdot- \sqrt{21} = \\ = 4- 2 \sqrt{21} + 2 \sqrt{21}-21 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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