Question
Answer
$$\dfrac{3-\sqrt{5}}{(\sqrt{5}+\sqrt{3})^2}=\dfrac{12-3\sqrt{15}-4\sqrt{5}+5\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3-\sqrt{5}}{(\sqrt{5}+\sqrt{3})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3-\sqrt{5}}{5+\sqrt{15}+\sqrt{15}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3-\sqrt{5}}{8+2\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3-\sqrt{5}}{8+2\sqrt{15}}\frac{8-2\sqrt{15}}{8-2\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{24-6\sqrt{15}-8\sqrt{5}+10\sqrt{3}}{64-16\sqrt{15}+16\sqrt{15}-60} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{24-6\sqrt{15}-8\sqrt{5}+10\sqrt{3}}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{12-3\sqrt{15}-4\sqrt{5}+5\sqrt{3}}{2}\end{aligned} $$ | |
| ① | $$ (\sqrt{5}+\sqrt{3})^2 = \left( \sqrt{5} + \sqrt{3} \right) \cdot \left( \sqrt{5} + \sqrt{3} \right) = 5 + \sqrt{15} + \sqrt{15} + 3 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 8- 2 \sqrt{15}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ \left( 3- \sqrt{5}\right) } \cdot \left( 8- 2 \sqrt{15}\right) = \color{blue}{3} \cdot8+\color{blue}{3} \cdot- 2 \sqrt{15}\color{blue}{- \sqrt{5}} \cdot8\color{blue}{- \sqrt{5}} \cdot- 2 \sqrt{15} = \\ = 24- 6 \sqrt{15}- 8 \sqrt{5} + 10 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 8 + 2 \sqrt{15}\right) } \cdot \left( 8- 2 \sqrt{15}\right) = \color{blue}{8} \cdot8+\color{blue}{8} \cdot- 2 \sqrt{15}+\color{blue}{ 2 \sqrt{15}} \cdot8+\color{blue}{ 2 \sqrt{15}} \cdot- 2 \sqrt{15} = \\ = 64- 16 \sqrt{15} + 16 \sqrt{15}-60 $$ |
| ⑤ | Simplify numerator and denominator |
| ⑥ | Divide both numerator and denominator by 2. |
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