Question
Answer
$$\dfrac{\sqrt{18}+11}{\sqrt{27}+3}=\dfrac{3\sqrt{6}-3\sqrt{2}+11\sqrt{3}-11}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{18}+11}{\sqrt{27}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{18}+11}{\sqrt{27}+3}\frac{\sqrt{27}-3}{\sqrt{27}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{9\sqrt{6}-9\sqrt{2}+33\sqrt{3}-33}{27-9\sqrt{3}+9\sqrt{3}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{9\sqrt{6}-9\sqrt{2}+33\sqrt{3}-33}{18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3\sqrt{6}-3\sqrt{2}+11\sqrt{3}-11}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{27}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{18} + 11\right) } \cdot \left( \sqrt{27}-3\right) = \color{blue}{ \sqrt{18}} \cdot \sqrt{27}+\color{blue}{ \sqrt{18}} \cdot-3+\color{blue}{11} \cdot \sqrt{27}+\color{blue}{11} \cdot-3 = \\ = 9 \sqrt{6}- 9 \sqrt{2} + 33 \sqrt{3}-33 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{27} + 3\right) } \cdot \left( \sqrt{27}-3\right) = \color{blue}{ \sqrt{27}} \cdot \sqrt{27}+\color{blue}{ \sqrt{27}} \cdot-3+\color{blue}{3} \cdot \sqrt{27}+\color{blue}{3} \cdot-3 = \\ = 27- 9 \sqrt{3} + 9 \sqrt{3}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
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