Question
Answer
$$\sqrt{12}-\sqrt{20}+\sqrt{27}-\sqrt{45}=5\sqrt{3}-5\sqrt{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\sqrt{12}-\sqrt{20}+\sqrt{27}-\sqrt{45}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2\sqrt{3}-2\sqrt{5}+3\sqrt{3}-3\sqrt{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}5\sqrt{3}-5\sqrt{5}\end{aligned} $$ | |
| ① | $$ \sqrt{12} =
\sqrt{ 2 ^2 \cdot 3 } =
\sqrt{ 2 ^2 } \, \sqrt{ 3 } =
2 \sqrt{ 3 }$$ |
| ② | $$ - \sqrt{20} =
- \sqrt{ 2 ^2 \cdot 5 } =
- \sqrt{ 2 ^2 } \, \sqrt{ 5 } =
- 2 \sqrt{ 5 }$$ |
| ③ | $$ \sqrt{27} =
\sqrt{ 3 ^2 \cdot 3 } =
\sqrt{ 3 ^2 } \, \sqrt{ 3 } =
3 \sqrt{ 3 }$$ |
| ④ | $$ - \sqrt{45} =
- \sqrt{ 3 ^2 \cdot 5 } =
- \sqrt{ 3 ^2 } \, \sqrt{ 5 } =
- 3 \sqrt{ 5 }$$ |
| ⑤ | Combine like terms |
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