Question
Answer
$$\dfrac{\sqrt{27}-\sqrt{5}}{\sqrt{15}-3}=\dfrac{3\sqrt{5}+2\sqrt{3}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{27}-\sqrt{5}}{\sqrt{15}-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{27}-\sqrt{5}}{\sqrt{15}-3}\frac{\sqrt{15}+3}{\sqrt{15}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{9\sqrt{5}+9\sqrt{3}-5\sqrt{3}-3\sqrt{5}}{15+3\sqrt{15}-3\sqrt{15}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{5}+4\sqrt{3}}{6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3\sqrt{5}+2\sqrt{3}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{15} + 3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{27}- \sqrt{5}\right) } \cdot \left( \sqrt{15} + 3\right) = \color{blue}{ \sqrt{27}} \cdot \sqrt{15}+\color{blue}{ \sqrt{27}} \cdot3\color{blue}{- \sqrt{5}} \cdot \sqrt{15}\color{blue}{- \sqrt{5}} \cdot3 = \\ = 9 \sqrt{5} + 9 \sqrt{3}- 5 \sqrt{3}- 3 \sqrt{5} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{15}-3\right) } \cdot \left( \sqrt{15} + 3\right) = \color{blue}{ \sqrt{15}} \cdot \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot3\color{blue}{-3} \cdot \sqrt{15}\color{blue}{-3} \cdot3 = \\ = 15 + 3 \sqrt{15}- 3 \sqrt{15}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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