Question
Answer
$$\dfrac{7\sqrt{6}+3}{6\sqrt{3}-2}=\dfrac{63\sqrt{2}+7\sqrt{6}+9\sqrt{3}+3}{52}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7\sqrt{6}+3}{6\sqrt{3}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7\sqrt{6}+3}{6\sqrt{3}-2}\frac{6\sqrt{3}+2}{6\sqrt{3}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{126\sqrt{2}+14\sqrt{6}+18\sqrt{3}+6}{108+12\sqrt{3}-12\sqrt{3}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{126\sqrt{2}+14\sqrt{6}+18\sqrt{3}+6}{104} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{63\sqrt{2}+7\sqrt{6}+9\sqrt{3}+3}{52}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6 \sqrt{3} + 2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7 \sqrt{6} + 3\right) } \cdot \left( 6 \sqrt{3} + 2\right) = \color{blue}{ 7 \sqrt{6}} \cdot 6 \sqrt{3}+\color{blue}{ 7 \sqrt{6}} \cdot2+\color{blue}{3} \cdot 6 \sqrt{3}+\color{blue}{3} \cdot2 = \\ = 126 \sqrt{2} + 14 \sqrt{6} + 18 \sqrt{3} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( 6 \sqrt{3}-2\right) } \cdot \left( 6 \sqrt{3} + 2\right) = \color{blue}{ 6 \sqrt{3}} \cdot 6 \sqrt{3}+\color{blue}{ 6 \sqrt{3}} \cdot2\color{blue}{-2} \cdot 6 \sqrt{3}\color{blue}{-2} \cdot2 = \\ = 108 + 12 \sqrt{3}- 12 \sqrt{3}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
This page was created using
Rationalize Denominator Calculator