Question
Answer
$$\dfrac{\sqrt{7}}{\sqrt{3}+2\sqrt{10}}=\dfrac{-\sqrt{21}+2\sqrt{70}}{37}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{7}}{\sqrt{3}+2\sqrt{10}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{7}}{\sqrt{3}+2\sqrt{10}}\frac{\sqrt{3}-2\sqrt{10}}{\sqrt{3}-2\sqrt{10}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{21}-2\sqrt{70}}{3-2\sqrt{30}+2\sqrt{30}-40} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{21}-2\sqrt{70}}{-37} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\sqrt{21}+2\sqrt{70}}{37}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}- 2 \sqrt{10}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{7} } \cdot \left( \sqrt{3}- 2 \sqrt{10}\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{3}+\color{blue}{ \sqrt{7}} \cdot- 2 \sqrt{10} = \\ = \sqrt{21}- 2 \sqrt{70} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + 2 \sqrt{10}\right) } \cdot \left( \sqrt{3}- 2 \sqrt{10}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot- 2 \sqrt{10}+\color{blue}{ 2 \sqrt{10}} \cdot \sqrt{3}+\color{blue}{ 2 \sqrt{10}} \cdot- 2 \sqrt{10} = \\ = 3- 2 \sqrt{30} + 2 \sqrt{30}-40 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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