Question
Answer
$$\dfrac{7}{-\sqrt{35}+10\sqrt{3}}=\dfrac{7\sqrt{35}+70\sqrt{3}}{265}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{-\sqrt{35}+10\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{-\sqrt{35}+10\sqrt{3}}\frac{-\sqrt{35}-10\sqrt{3}}{-\sqrt{35}-10\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-7\sqrt{35}-70\sqrt{3}}{35+10\sqrt{105}-10\sqrt{105}-300} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-7\sqrt{35}-70\sqrt{3}}{-265} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7\sqrt{35}+70\sqrt{3}}{265}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ - \sqrt{35}- 10 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( - \sqrt{35}- 10 \sqrt{3}\right) = \color{blue}{7} \cdot- \sqrt{35}+\color{blue}{7} \cdot- 10 \sqrt{3} = \\ = - 7 \sqrt{35}- 70 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( - \sqrt{35} + 10 \sqrt{3}\right) } \cdot \left( - \sqrt{35}- 10 \sqrt{3}\right) = \color{blue}{- \sqrt{35}} \cdot- \sqrt{35}\color{blue}{- \sqrt{35}} \cdot- 10 \sqrt{3}+\color{blue}{ 10 \sqrt{3}} \cdot- \sqrt{35}+\color{blue}{ 10 \sqrt{3}} \cdot- 10 \sqrt{3} = \\ = 35 + 10 \sqrt{105}- 10 \sqrt{105}-300 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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