Question
Answer
$$\dfrac{2\sqrt{3}+1}{4\sqrt{2}-3}=\dfrac{8\sqrt{6}+6\sqrt{3}+4\sqrt{2}+3}{23}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{3}+1}{4\sqrt{2}-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{3}+1}{4\sqrt{2}-3}\frac{4\sqrt{2}+3}{4\sqrt{2}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8\sqrt{6}+6\sqrt{3}+4\sqrt{2}+3}{32+12\sqrt{2}-12\sqrt{2}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8\sqrt{6}+6\sqrt{3}+4\sqrt{2}+3}{23}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 \sqrt{2} + 3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 \sqrt{3} + 1\right) } \cdot \left( 4 \sqrt{2} + 3\right) = \color{blue}{ 2 \sqrt{3}} \cdot 4 \sqrt{2}+\color{blue}{ 2 \sqrt{3}} \cdot3+\color{blue}{1} \cdot 4 \sqrt{2}+\color{blue}{1} \cdot3 = \\ = 8 \sqrt{6} + 6 \sqrt{3} + 4 \sqrt{2} + 3 $$ Simplify denominator. $$ \color{blue}{ \left( 4 \sqrt{2}-3\right) } \cdot \left( 4 \sqrt{2} + 3\right) = \color{blue}{ 4 \sqrt{2}} \cdot 4 \sqrt{2}+\color{blue}{ 4 \sqrt{2}} \cdot3\color{blue}{-3} \cdot 4 \sqrt{2}\color{blue}{-3} \cdot3 = \\ = 32 + 12 \sqrt{2}- 12 \sqrt{2}-9 $$ |
| ③ | Simplify numerator and denominator |
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