Answer
$$\frac{7-5\sqrt{3}}{7+5\sqrt{3}}=\frac{-62+35\sqrt{3}}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7-5\sqrt{3}}{7+5\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7-5\sqrt{3}}{7+5\sqrt{3}}\frac{7-5\sqrt{3}}{7-5\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{49-35\sqrt{3}-35\sqrt{3}+75}{49-35\sqrt{3}+35\sqrt{3}-75} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{124-70\sqrt{3}}{-26} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{62-35\sqrt{3}}{-13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-62+35\sqrt{3}}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- 5 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7- 5 \sqrt{3}\right) } \cdot \left( 7- 5 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 5 \sqrt{3}\color{blue}{- 5 \sqrt{3}} \cdot7\color{blue}{- 5 \sqrt{3}} \cdot- 5 \sqrt{3} = \\ = 49- 35 \sqrt{3}- 35 \sqrt{3} + 75 $$ Simplify denominator. $$ \color{blue}{ \left( 7 + 5 \sqrt{3}\right) } \cdot \left( 7- 5 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 5 \sqrt{3}+\color{blue}{ 5 \sqrt{3}} \cdot7+\color{blue}{ 5 \sqrt{3}} \cdot- 5 \sqrt{3} = \\ = 49- 35 \sqrt{3} + 35 \sqrt{3}-75 $$ |
| 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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