Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{6+\sqrt{10}}{7-\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6+\sqrt{10}}{7-\sqrt{3}}\frac{7+\sqrt{3}}{7+\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{42+6\sqrt{3}+7\sqrt{10}+\sqrt{30}}{49+7\sqrt{3}-7\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{42+6\sqrt{3}+7\sqrt{10}+\sqrt{30}}{46}\end{aligned} $$ | |
① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7 + \sqrt{3}} $$. |
② | Multiply in a numerator. $$ \color{blue}{ \left( 6 + \sqrt{10}\right) } \cdot \left( 7 + \sqrt{3}\right) = \color{blue}{6} \cdot7+\color{blue}{6} \cdot \sqrt{3}+\color{blue}{ \sqrt{10}} \cdot7+\color{blue}{ \sqrt{10}} \cdot \sqrt{3} = \\ = 42 + 6 \sqrt{3} + 7 \sqrt{10} + \sqrt{30} $$ Simplify denominator. $$ \color{blue}{ \left( 7- \sqrt{3}\right) } \cdot \left( 7 + \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot \sqrt{3}\color{blue}{- \sqrt{3}} \cdot7\color{blue}{- \sqrt{3}} \cdot \sqrt{3} = \\ = 49 + 7 \sqrt{3}- 7 \sqrt{3}-3 $$ |
③ | Simplify numerator and denominator |