Answer
$$\frac{7}{2+\sqrt{19}}=\frac{-14+7\sqrt{19}}{15}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{2+\sqrt{19}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{2+\sqrt{19}}\frac{2-\sqrt{19}}{2-\sqrt{19}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14-7\sqrt{19}}{4-2\sqrt{19}+2\sqrt{19}-19} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14-7\sqrt{19}}{-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-14+7\sqrt{19}}{15}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- \sqrt{19}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( 2- \sqrt{19}\right) = \color{blue}{7} \cdot2+\color{blue}{7} \cdot- \sqrt{19} = \\ = 14- 7 \sqrt{19} $$ Simplify denominator. $$ \color{blue}{ \left( 2 + \sqrt{19}\right) } \cdot \left( 2- \sqrt{19}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- \sqrt{19}+\color{blue}{ \sqrt{19}} \cdot2+\color{blue}{ \sqrt{19}} \cdot- \sqrt{19} = \\ = 4- 2 \sqrt{19} + 2 \sqrt{19}-19 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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