Question
Answer
$$\sqrt{3}\cdot(\sqrt{27}-\sqrt{15})+\sqrt{5}=9-2\sqrt{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\sqrt{3}\cdot(\sqrt{27}-\sqrt{15})+\sqrt{5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\sqrt{3}\cdot(3\sqrt{3}-\sqrt{15})+\sqrt{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}9-3\sqrt{5}+\sqrt{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}9-2\sqrt{5}\end{aligned} $$ | |
| ① | $$ \sqrt{27} =
\sqrt{ 3 ^2 \cdot 3 } =
\sqrt{ 3 ^2 } \, \sqrt{ 3 } =
3 \sqrt{ 3 }$$ |
| ② | $$ \color{blue}{ \sqrt{3} } \cdot \left( 3 \sqrt{3}- \sqrt{15}\right) = \color{blue}{ \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{15} = \\ = 9- 3 \sqrt{5} $$ |
| ③ | Combine like terms |
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