Question
Answer
$$\dfrac{\sqrt{12}-\sqrt{7}}{\sqrt{12}+\sqrt{7}}=\dfrac{19-4\sqrt{21}}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{12}-\sqrt{7}}{\sqrt{12}+\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{12}-\sqrt{7}}{\sqrt{12}+\sqrt{7}}\frac{\sqrt{12}-\sqrt{7}}{\sqrt{12}-\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12-2\sqrt{21}-2\sqrt{21}+7}{12-2\sqrt{21}+2\sqrt{21}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{19-4\sqrt{21}}{5}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{12}- \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{12}- \sqrt{7}\right) } \cdot \left( \sqrt{12}- \sqrt{7}\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot- \sqrt{7}\color{blue}{- \sqrt{7}} \cdot \sqrt{12}\color{blue}{- \sqrt{7}} \cdot- \sqrt{7} = \\ = 12- 2 \sqrt{21}- 2 \sqrt{21} + 7 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{12} + \sqrt{7}\right) } \cdot \left( \sqrt{12}- \sqrt{7}\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot- \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot \sqrt{12}+\color{blue}{ \sqrt{7}} \cdot- \sqrt{7} = \\ = 12- 2 \sqrt{21} + 2 \sqrt{21}-7 $$ |
| ③ | Simplify numerator and denominator |
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