Question
Answer
$$\dfrac{7\sqrt{2}+1}{6\sqrt{3}-1}=\dfrac{42\sqrt{6}+7\sqrt{2}+6\sqrt{3}+1}{107}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7\sqrt{2}+1}{6\sqrt{3}-1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7\sqrt{2}+1}{6\sqrt{3}-1}\frac{6\sqrt{3}+1}{6\sqrt{3}+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{42\sqrt{6}+7\sqrt{2}+6\sqrt{3}+1}{108+6\sqrt{3}-6\sqrt{3}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{42\sqrt{6}+7\sqrt{2}+6\sqrt{3}+1}{107}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6 \sqrt{3} + 1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7 \sqrt{2} + 1\right) } \cdot \left( 6 \sqrt{3} + 1\right) = \color{blue}{ 7 \sqrt{2}} \cdot 6 \sqrt{3}+\color{blue}{ 7 \sqrt{2}} \cdot1+\color{blue}{1} \cdot 6 \sqrt{3}+\color{blue}{1} \cdot1 = \\ = 42 \sqrt{6} + 7 \sqrt{2} + 6 \sqrt{3} + 1 $$ Simplify denominator. $$ \color{blue}{ \left( 6 \sqrt{3}-1\right) } \cdot \left( 6 \sqrt{3} + 1\right) = \color{blue}{ 6 \sqrt{3}} \cdot 6 \sqrt{3}+\color{blue}{ 6 \sqrt{3}} \cdot1\color{blue}{-1} \cdot 6 \sqrt{3}\color{blue}{-1} \cdot1 = \\ = 108 + 6 \sqrt{3}- 6 \sqrt{3}-1 $$ |
| ③ | Simplify numerator and denominator |
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