Question
Answer
$$\dfrac{\sqrt{72}}{5\sqrt{72}+3\sqrt{288}-2\sqrt{648}}=\dfrac{1}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{72}}{5\sqrt{72}+3\sqrt{288}-2\sqrt{648}}& \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}} } }}}\frac{ \sqrt{ 36 \cdot 2 } }{ 30\sqrt{2}+36\sqrt{2}-36\sqrt{2} } \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{ \sqrt{ 36 } \cdot \sqrt{ 2 } }{ 30\sqrt{2}+36\sqrt{2}-36\sqrt{2} } \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} \htmlClass{explanationCircle explanationCircle12}{\textcircled {12}} } }}}\frac{6\sqrt{2}}{30\sqrt{2}+36\sqrt{2}-36\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle13}{\textcircled {13}} } }}}\frac{6\sqrt{2}}{30\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle14}{\textcircled {14}} } }}}\frac{6\sqrt{2}}{30\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle15}{\textcircled {15}} } }}}\frac{12}{60} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle16}{\textcircled {16}} } }}} \frac{ 12 : \color{orangered}{ 12 } }{ 60 : \color{orangered}{ 12 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{1}{5}\end{aligned} $$ | |
| ① | Factor out the largest perfect square of 72. ( in this example we factored out $ 36 $ ) |
| ② | $$ 5 \sqrt{72} =
5 \sqrt{ 6 ^2 \cdot 2 } =
5 \sqrt{ 6 ^2 } \, \sqrt{ 2 } =
5 \cdot 6 \sqrt{ 2 } =
30 \sqrt{ 2 } $$ |
| ③ | $$ 3 \sqrt{288} =
3 \sqrt{ 12 ^2 \cdot 2 } =
3 \sqrt{ 12 ^2 } \, \sqrt{ 2 } =
3 \cdot 12 \sqrt{ 2 } =
36 \sqrt{ 2 } $$ |
| ④ | $$ - 2 \sqrt{648} =
-2 \sqrt{ 18 ^2 \cdot 2 } =
-2 \sqrt{ 18 ^2 } \, \sqrt{ 2 } =
-2 \cdot 18 \sqrt{ 2 } =
-36 \sqrt{ 2 } $$ |
| ⑤ | Rewrite $ \sqrt{ 36 \cdot 2 } $ as the product of two radicals. |
| ⑥ | $$ 5 \sqrt{72} =
5 \sqrt{ 6 ^2 \cdot 2 } =
5 \sqrt{ 6 ^2 } \, \sqrt{ 2 } =
5 \cdot 6 \sqrt{ 2 } =
30 \sqrt{ 2 } $$ |
| ⑦ | $$ 3 \sqrt{288} =
3 \sqrt{ 12 ^2 \cdot 2 } =
3 \sqrt{ 12 ^2 } \, \sqrt{ 2 } =
3 \cdot 12 \sqrt{ 2 } =
36 \sqrt{ 2 } $$ |
| ⑧ | $$ - 2 \sqrt{648} =
-2 \sqrt{ 18 ^2 \cdot 2 } =
-2 \sqrt{ 18 ^2 } \, \sqrt{ 2 } =
-2 \cdot 18 \sqrt{ 2 } =
-36 \sqrt{ 2 } $$ |
| ⑨ | The square root of $ 36 $ is $ 6 $. |
| ⑩ | $$ 5 \sqrt{72} =
5 \sqrt{ 6 ^2 \cdot 2 } =
5 \sqrt{ 6 ^2 } \, \sqrt{ 2 } =
5 \cdot 6 \sqrt{ 2 } =
30 \sqrt{ 2 } $$ |
| ⑪ | $$ 3 \sqrt{288} =
3 \sqrt{ 12 ^2 \cdot 2 } =
3 \sqrt{ 12 ^2 } \, \sqrt{ 2 } =
3 \cdot 12 \sqrt{ 2 } =
36 \sqrt{ 2 } $$ |
| ⑫ | $$ - 2 \sqrt{648} =
-2 \sqrt{ 18 ^2 \cdot 2 } =
-2 \sqrt{ 18 ^2 } \, \sqrt{ 2 } =
-2 \cdot 18 \sqrt{ 2 } =
-36 \sqrt{ 2 } $$ |
| ⑬ | Simplify numerator and denominator |
| ⑭ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}} $$. |
| ⑮ | Multiply in a numerator. $$ \color{blue}{ 6 \sqrt{2} } \cdot \sqrt{2} = 12 $$ Simplify denominator. $$ \color{blue}{ 30 \sqrt{2} } \cdot \sqrt{2} = 60 $$ |
| ⑯ | Divide both the top and bottom numbers by $ \color{orangered}{ 12 } $. |
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