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