Question
Answer
$$\dfrac{3\sqrt{11}}{\sqrt{33}}+\sqrt{48}=5\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{11}}{\sqrt{33}}+\sqrt{48}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{11}}{\sqrt{33}}+4\sqrt{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{33\sqrt{3}}{33}+4\sqrt{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{3}}{1}+4\sqrt{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\sqrt{3}+4\sqrt{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}5\sqrt{3}\end{aligned} $$ | |
| ① | $$ \sqrt{48} =
\sqrt{ 4 ^2 \cdot 3 } =
\sqrt{ 4 ^2 } \, \sqrt{ 3 } =
4 \sqrt{ 3 }$$ |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 \sqrt{11} } \cdot \sqrt{33} = 33 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \sqrt{33} } \cdot \sqrt{33} = 33 $$ |
| ③ | Divide both numerator and denominator by 33. |
| ④ | Remove 1 from denominator. |
| ⑤ | Combine like terms |
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