Question
Answer
$$\dfrac{\sqrt{3}}{3}+2\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}=\dfrac{5\sqrt{3}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{3}+2\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{3}+6\sqrt{3}}{6}-\frac{\sqrt{3}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8\sqrt{3}}{6}-\frac{\sqrt{3}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8\sqrt{3}-3\sqrt{3}}{6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{5\sqrt{3}}{6}\end{aligned} $$ | |
| ① | $$ \frac{\sqrt{3}}{3}+2\frac{\sqrt{3}}{2}
= \frac{\sqrt{3}}{3} \cdot \color{blue}{\frac{ 2 }{ 2}} + \frac{2\sqrt{3}}{2} \cdot \color{blue}{\frac{ 3 }{ 3}}
= \frac{2\sqrt{3}+6\sqrt{3}}{6} $$ |
| ② | Simplify numerator and denominator |
| ③ | $$ \frac{8\sqrt{3}}{6}-\frac{\sqrt{3}}{2}
= \frac{8\sqrt{3}}{6} \cdot \color{blue}{\frac{ 1 }{ 1}} - \frac{\sqrt{3}}{2} \cdot \color{blue}{\frac{ 3 }{ 3}}
= \frac{8\sqrt{3}-3\sqrt{3}}{6} $$ |
| ④ | Simplify numerator and denominator |
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