Answer
$$\frac{2-\sqrt{7}}{\sqrt{12}+3}=\frac{4\sqrt{3}-6-2\sqrt{21}+3\sqrt{7}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-\sqrt{7}}{\sqrt{12}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-\sqrt{7}}{\sqrt{12}+3}\frac{\sqrt{12}-3}{\sqrt{12}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{3}-6-2\sqrt{21}+3\sqrt{7}}{12-6\sqrt{3}+6\sqrt{3}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{3}-6-2\sqrt{21}+3\sqrt{7}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{12}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2- \sqrt{7}\right) } \cdot \left( \sqrt{12}-3\right) = \color{blue}{2} \cdot \sqrt{12}+\color{blue}{2} \cdot-3\color{blue}{- \sqrt{7}} \cdot \sqrt{12}\color{blue}{- \sqrt{7}} \cdot-3 = \\ = 4 \sqrt{3}-6- 2 \sqrt{21} + 3 \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{12} + 3\right) } \cdot \left( \sqrt{12}-3\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot-3+\color{blue}{3} \cdot \sqrt{12}+\color{blue}{3} \cdot-3 = \\ = 12- 6 \sqrt{3} + 6 \sqrt{3}-9 $$ |
| ③ | Simplify numerator and denominator |
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