Question
Answer
$$\dfrac{4\sqrt{72}+11\sqrt{63}-2\sqrt{28}}{1}=24\sqrt{2}+29\sqrt{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4\sqrt{72}+11\sqrt{63}-2\sqrt{28}}{1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4\sqrt{72}+11\sqrt{63}-2\sqrt{28} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}24\sqrt{2}+33\sqrt{7}-4\sqrt{7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}24\sqrt{2}+29\sqrt{7}\end{aligned} $$ | |
| ① | Remove 1 from denominator. |
| ② | $$ 4 \sqrt{72} =
4 \sqrt{ 6 ^2 \cdot 2 } =
4 \sqrt{ 6 ^2 } \, \sqrt{ 2 } =
4 \cdot 6 \sqrt{ 2 } =
24 \sqrt{ 2 } $$ |
| ③ | $$ 11 \sqrt{63} =
11 \sqrt{ 3 ^2 \cdot 7 } =
11 \sqrt{ 3 ^2 } \, \sqrt{ 7 } =
11 \cdot 3 \sqrt{ 7 } =
33 \sqrt{ 7 } $$ |
| ④ | $$ - 2 \sqrt{28} =
-2 \sqrt{ 2 ^2 \cdot 7 } =
-2 \sqrt{ 2 ^2 } \, \sqrt{ 7 } =
-2 \cdot 2 \sqrt{ 7 } =
-4 \sqrt{ 7 } $$ |
| ⑤ | Combine like terms |
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