Question
Answer
$$2\sqrt{11}+3\cdot\dfrac{\sqrt{44}}{4}-\sqrt{99}=\dfrac{\sqrt{11}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2\sqrt{11}+3 \cdot \frac{\sqrt{44}}{4}-\sqrt{99}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2\sqrt{11}+3 \cdot \frac{\sqrt{44}}{4}-3\sqrt{11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2\sqrt{11}+3 \cdot \frac{\sqrt{11}}{2}-3\sqrt{11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{11}+3\sqrt{11}}{2}-3\sqrt{11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7\sqrt{11}}{2}-3\sqrt{11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{7\sqrt{11}-6\sqrt{11}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{\sqrt{11}}{2}\end{aligned} $$ | |
| ① | $$ \sqrt{99} =
\sqrt{ 3 ^2 \cdot 11 } =
\sqrt{ 3 ^2 } \, \sqrt{ 11 } =
3 \sqrt{ 11 }$$ |
| ② | Divide both numerator and denominator by 2. |
| ③ | $$ 2\sqrt{11}+3 \cdot \frac{\sqrt{11}}{2}
= 2\sqrt{11} \cdot \color{blue}{\frac{ 2 }{ 2}} + \frac{3\sqrt{11}}{2} \cdot \color{blue}{\frac{ 1 }{ 1}}
= \frac{4\sqrt{11}+3\sqrt{11}}{2} $$ |
| ④ | Simplify numerator and denominator |
| ⑤ | $$ \frac{7\sqrt{11}}{2}-3\sqrt{11}
= \frac{7\sqrt{11}}{2} \cdot \color{blue}{\frac{ 1 }{ 1}} - 3\sqrt{11} \cdot \color{blue}{\frac{ 2 }{ 2}}
= \frac{7\sqrt{11}-6\sqrt{11}}{2} $$ |
| ⑥ | Simplify numerator and denominator |
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