Question
Answer
$$\dfrac{3\sqrt{5}+2\sqrt{7}}{7\sqrt{8}+8\sqrt{3}}=\dfrac{21\sqrt{10}-12\sqrt{15}+14\sqrt{14}-8\sqrt{21}}{100}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{5}+2\sqrt{7}}{7\sqrt{8}+8\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{5}+2\sqrt{7}}{7\sqrt{8}+8\sqrt{3}}\frac{7\sqrt{8}-8\sqrt{3}}{7\sqrt{8}-8\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{42\sqrt{10}-24\sqrt{15}+28\sqrt{14}-16\sqrt{21}}{392-112\sqrt{6}+112\sqrt{6}-192} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{42\sqrt{10}-24\sqrt{15}+28\sqrt{14}-16\sqrt{21}}{200} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{21\sqrt{10}-12\sqrt{15}+14\sqrt{14}-8\sqrt{21}}{100}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7 \sqrt{8}- 8 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 \sqrt{5} + 2 \sqrt{7}\right) } \cdot \left( 7 \sqrt{8}- 8 \sqrt{3}\right) = \color{blue}{ 3 \sqrt{5}} \cdot 7 \sqrt{8}+\color{blue}{ 3 \sqrt{5}} \cdot- 8 \sqrt{3}+\color{blue}{ 2 \sqrt{7}} \cdot 7 \sqrt{8}+\color{blue}{ 2 \sqrt{7}} \cdot- 8 \sqrt{3} = \\ = 42 \sqrt{10}- 24 \sqrt{15} + 28 \sqrt{14}- 16 \sqrt{21} $$ Simplify denominator. $$ \color{blue}{ \left( 7 \sqrt{8} + 8 \sqrt{3}\right) } \cdot \left( 7 \sqrt{8}- 8 \sqrt{3}\right) = \color{blue}{ 7 \sqrt{8}} \cdot 7 \sqrt{8}+\color{blue}{ 7 \sqrt{8}} \cdot- 8 \sqrt{3}+\color{blue}{ 8 \sqrt{3}} \cdot 7 \sqrt{8}+\color{blue}{ 8 \sqrt{3}} \cdot- 8 \sqrt{3} = \\ = 392- 112 \sqrt{6} + 112 \sqrt{6}-192 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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