Question
Answer
$$\dfrac{\sqrt{3}-\sqrt{8}}{\sqrt{8}+\sqrt{11}}=\dfrac{-2\sqrt{6}+\sqrt{33}+8-2\sqrt{22}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}-\sqrt{8}}{\sqrt{8}+\sqrt{11}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}-\sqrt{8}}{\sqrt{8}+\sqrt{11}}\frac{\sqrt{8}-\sqrt{11}}{\sqrt{8}-\sqrt{11}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{6}-\sqrt{33}-8+2\sqrt{22}}{8-2\sqrt{22}+2\sqrt{22}-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{6}-\sqrt{33}-8+2\sqrt{22}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2\sqrt{6}+\sqrt{33}+8-2\sqrt{22}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{8}- \sqrt{11}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3}- \sqrt{8}\right) } \cdot \left( \sqrt{8}- \sqrt{11}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{8}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{11}\color{blue}{- \sqrt{8}} \cdot \sqrt{8}\color{blue}{- \sqrt{8}} \cdot- \sqrt{11} = \\ = 2 \sqrt{6}- \sqrt{33}-8 + 2 \sqrt{22} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{8} + \sqrt{11}\right) } \cdot \left( \sqrt{8}- \sqrt{11}\right) = \color{blue}{ \sqrt{8}} \cdot \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot- \sqrt{11}+\color{blue}{ \sqrt{11}} \cdot \sqrt{8}+\color{blue}{ \sqrt{11}} \cdot- \sqrt{11} = \\ = 8- 2 \sqrt{22} + 2 \sqrt{22}-11 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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