Answer
$$\frac{\sqrt{3}-4}{\sqrt{7}+2}=\frac{\sqrt{21}-2\sqrt{3}-4\sqrt{7}+8}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}-4}{\sqrt{7}+2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}-4}{\sqrt{7}+2}\frac{\sqrt{7}-2}{\sqrt{7}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{21}-2\sqrt{3}-4\sqrt{7}+8}{7-2\sqrt{7}+2\sqrt{7}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{21}-2\sqrt{3}-4\sqrt{7}+8}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7}-2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3}-4\right) } \cdot \left( \sqrt{7}-2\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{7}+\color{blue}{ \sqrt{3}} \cdot-2\color{blue}{-4} \cdot \sqrt{7}\color{blue}{-4} \cdot-2 = \\ = \sqrt{21}- 2 \sqrt{3}- 4 \sqrt{7} + 8 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7} + 2\right) } \cdot \left( \sqrt{7}-2\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot-2+\color{blue}{2} \cdot \sqrt{7}+\color{blue}{2} \cdot-2 = \\ = 7- 2 \sqrt{7} + 2 \sqrt{7}-4 $$ |
| ③ | Simplify numerator and denominator |
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