Question
Answer
$$\dfrac{2-\sqrt{10}}{7+\sqrt{10}}=\dfrac{8-3\sqrt{10}}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-\sqrt{10}}{7+\sqrt{10}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-\sqrt{10}}{7+\sqrt{10}}\frac{7-\sqrt{10}}{7-\sqrt{10}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14-2\sqrt{10}-7\sqrt{10}+10}{49-7\sqrt{10}+7\sqrt{10}-10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{24-9\sqrt{10}}{39} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{8-3\sqrt{10}}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- \sqrt{10}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2- \sqrt{10}\right) } \cdot \left( 7- \sqrt{10}\right) = \color{blue}{2} \cdot7+\color{blue}{2} \cdot- \sqrt{10}\color{blue}{- \sqrt{10}} \cdot7\color{blue}{- \sqrt{10}} \cdot- \sqrt{10} = \\ = 14- 2 \sqrt{10}- 7 \sqrt{10} + 10 $$ Simplify denominator. $$ \color{blue}{ \left( 7 + \sqrt{10}\right) } \cdot \left( 7- \sqrt{10}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- \sqrt{10}+\color{blue}{ \sqrt{10}} \cdot7+\color{blue}{ \sqrt{10}} \cdot- \sqrt{10} = \\ = 49- 7 \sqrt{10} + 7 \sqrt{10}-10 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
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