Question
Answer
$$\dfrac{4\sqrt{10}-2\sqrt{3}}{\sqrt{10}+\sqrt{3}}=\dfrac{46-6\sqrt{30}}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4\sqrt{10}-2\sqrt{3}}{\sqrt{10}+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4\sqrt{10}-2\sqrt{3}}{\sqrt{10}+\sqrt{3}}\frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{40-4\sqrt{30}-2\sqrt{30}+6}{10-\sqrt{30}+\sqrt{30}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{46-6\sqrt{30}}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{10}- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 4 \sqrt{10}- 2 \sqrt{3}\right) } \cdot \left( \sqrt{10}- \sqrt{3}\right) = \color{blue}{ 4 \sqrt{10}} \cdot \sqrt{10}+\color{blue}{ 4 \sqrt{10}} \cdot- \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot \sqrt{10}\color{blue}{- 2 \sqrt{3}} \cdot- \sqrt{3} = \\ = 40- 4 \sqrt{30}- 2 \sqrt{30} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{10} + \sqrt{3}\right) } \cdot \left( \sqrt{10}- \sqrt{3}\right) = \color{blue}{ \sqrt{10}} \cdot \sqrt{10}+\color{blue}{ \sqrt{10}} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot \sqrt{10}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 10- \sqrt{30} + \sqrt{30}-3 $$ |
| ③ | Simplify numerator and denominator |
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