Question
Answer
$$\dfrac{7\sqrt{5}-4}{6+5\sqrt{3}}=\dfrac{-42\sqrt{5}+35\sqrt{15}+24-20\sqrt{3}}{39}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7\sqrt{5}-4}{6+5\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7\sqrt{5}-4}{6+5\sqrt{3}}\frac{6-5\sqrt{3}}{6-5\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{42\sqrt{5}-35\sqrt{15}-24+20\sqrt{3}}{36-30\sqrt{3}+30\sqrt{3}-75} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{42\sqrt{5}-35\sqrt{15}-24+20\sqrt{3}}{-39} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-42\sqrt{5}+35\sqrt{15}+24-20\sqrt{3}}{39}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- 5 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7 \sqrt{5}-4\right) } \cdot \left( 6- 5 \sqrt{3}\right) = \color{blue}{ 7 \sqrt{5}} \cdot6+\color{blue}{ 7 \sqrt{5}} \cdot- 5 \sqrt{3}\color{blue}{-4} \cdot6\color{blue}{-4} \cdot- 5 \sqrt{3} = \\ = 42 \sqrt{5}- 35 \sqrt{15}-24 + 20 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 6 + 5 \sqrt{3}\right) } \cdot \left( 6- 5 \sqrt{3}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- 5 \sqrt{3}+\color{blue}{ 5 \sqrt{3}} \cdot6+\color{blue}{ 5 \sqrt{3}} \cdot- 5 \sqrt{3} = \\ = 36- 30 \sqrt{3} + 30 \sqrt{3}-75 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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