Question
Answer
$$\dfrac{1}{\sqrt{32}+\sqrt{50}+\sqrt{72}}=\dfrac{\sqrt{2}}{30}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{32}+\sqrt{50}+\sqrt{72}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1}{4\sqrt{2}+5\sqrt{2}+6\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{1}{15\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{1}{15\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{\sqrt{2}}{30}\end{aligned} $$ | |
| ① | $$ \sqrt{32} =
\sqrt{ 4 ^2 \cdot 2 } =
\sqrt{ 4 ^2 } \, \sqrt{ 2 } =
4 \sqrt{ 2 }$$ |
| ② | $$ \sqrt{50} =
\sqrt{ 5 ^2 \cdot 2 } =
\sqrt{ 5 ^2 } \, \sqrt{ 2 } =
5 \sqrt{ 2 }$$ |
| ③ | $$ \sqrt{72} =
\sqrt{ 6 ^2 \cdot 2 } =
\sqrt{ 6 ^2 } \, \sqrt{ 2 } =
6 \sqrt{ 2 }$$ |
| ④ | Simplify numerator and denominator |
| ⑤ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}} $$. |
| ⑥ | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \sqrt{2} = \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ 15 \sqrt{2} } \cdot \sqrt{2} = 30 $$ |
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