Question
Answer
$$\dfrac{\sqrt{7}+3}{2\sqrt{7}-5}=\dfrac{29+11\sqrt{7}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{7}+3}{2\sqrt{7}-5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{7}+3}{2\sqrt{7}-5}\frac{2\sqrt{7}+5}{2\sqrt{7}+5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14+5\sqrt{7}+6\sqrt{7}+15}{28+10\sqrt{7}-10\sqrt{7}-25} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{29+11\sqrt{7}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{7} + 5} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{7} + 3\right) } \cdot \left( 2 \sqrt{7} + 5\right) = \color{blue}{ \sqrt{7}} \cdot 2 \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot5+\color{blue}{3} \cdot 2 \sqrt{7}+\color{blue}{3} \cdot5 = \\ = 14 + 5 \sqrt{7} + 6 \sqrt{7} + 15 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{7}-5\right) } \cdot \left( 2 \sqrt{7} + 5\right) = \color{blue}{ 2 \sqrt{7}} \cdot 2 \sqrt{7}+\color{blue}{ 2 \sqrt{7}} \cdot5\color{blue}{-5} \cdot 2 \sqrt{7}\color{blue}{-5} \cdot5 = \\ = 28 + 10 \sqrt{7}- 10 \sqrt{7}-25 $$ |
| ③ | Simplify numerator and denominator |
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