Question
Answer
$$\dfrac{7+\sqrt{63}}{5+\sqrt{63}}=\dfrac{14+3\sqrt{7}}{19}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7+\sqrt{63}}{5+\sqrt{63}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7+\sqrt{63}}{5+\sqrt{63}}\frac{5-\sqrt{63}}{5-\sqrt{63}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{35-21\sqrt{7}+15\sqrt{7}-63}{25-15\sqrt{7}+15\sqrt{7}-63} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-28-6\sqrt{7}}{-38} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-14-3\sqrt{7}}{-19} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{14+3\sqrt{7}}{19}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 5- \sqrt{63}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7 + \sqrt{63}\right) } \cdot \left( 5- \sqrt{63}\right) = \color{blue}{7} \cdot5+\color{blue}{7} \cdot- \sqrt{63}+\color{blue}{ \sqrt{63}} \cdot5+\color{blue}{ \sqrt{63}} \cdot- \sqrt{63} = \\ = 35- 21 \sqrt{7} + 15 \sqrt{7}-63 $$ Simplify denominator. $$ \color{blue}{ \left( 5 + \sqrt{63}\right) } \cdot \left( 5- \sqrt{63}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot- \sqrt{63}+\color{blue}{ \sqrt{63}} \cdot5+\color{blue}{ \sqrt{63}} \cdot- \sqrt{63} = \\ = 25- 15 \sqrt{7} + 15 \sqrt{7}-63 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
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