Answer
$$\frac{4-\sqrt{7}\cdot(2\sqrt{7}-3)}{2\sqrt{2}+3(2\sqrt{2}-3)}=\frac{-80\sqrt{2}-90+24\sqrt{14}+27\sqrt{7}}{47}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4-\sqrt{7}\cdot(2\sqrt{7}-3)}{2\sqrt{2}+3(2\sqrt{2}-3)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\frac{4-14+3\sqrt{7}}{1}}{\frac{2\sqrt{2}+6\sqrt{2}-9}{1}} \xlongequal{ } \\[1 em] & \xlongequal{ }(4-14+3\sqrt{7})\cdot\frac{1}{2\sqrt{2}+6\sqrt{2}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4-14+3\sqrt{7}}{2\sqrt{2}+6\sqrt{2}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-10+3\sqrt{7}}{8\sqrt{2}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-10+3\sqrt{7}}{8\sqrt{2}-9}\frac{8\sqrt{2}+9}{8\sqrt{2}+9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-80\sqrt{2}-90+24\sqrt{14}+27\sqrt{7}}{128+72\sqrt{2}-72\sqrt{2}-81} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{-80\sqrt{2}-90+24\sqrt{14}+27\sqrt{7}}{47}\end{aligned} $$ | |
| ① | $$ 4-\sqrt{7}\cdot(2\sqrt{7}-3)
= 4 \cdot \color{blue}{\frac{ 1 }{ 1}} - 14-3\sqrt{7} \cdot \color{blue}{\frac{ 1 }{ 1}}
= \frac{4-14+3\sqrt{7}}{1} $$ |
| ② | $$ 2\sqrt{2}+3(2\sqrt{2}-3)
= 2\sqrt{2} \cdot \color{blue}{\frac{ 1 }{ 1}} + 6\sqrt{2}-9 \cdot \color{blue}{\frac{ 1 }{ 1}}
= \frac{2\sqrt{2}+6\sqrt{2}-9}{1} $$ |
| ③ | $$ \color{blue}{ \left( 4-14 + 3 \sqrt{7}\right) } \cdot 1 = \color{blue}{4} \cdot1\color{blue}{-14} \cdot1+\color{blue}{ 3 \sqrt{7}} \cdot1 = \\ = 4-14 + 3 \sqrt{7} $$$$ \color{blue}{ 1 } \cdot \left( 2 \sqrt{2} + 6 \sqrt{2}-9\right) = \color{blue}{1} \cdot 2 \sqrt{2}+\color{blue}{1} \cdot 6 \sqrt{2}+\color{blue}{1} \cdot-9 = \\ = 2 \sqrt{2} + 6 \sqrt{2}-9 $$ |
| ④ | Simplify numerator and denominator |
| ⑤ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 8 \sqrt{2} + 9} $$. |
| ⑥ | Multiply in a numerator. $$ \color{blue}{ \left( -10 + 3 \sqrt{7}\right) } \cdot \left( 8 \sqrt{2} + 9\right) = \color{blue}{-10} \cdot 8 \sqrt{2}\color{blue}{-10} \cdot9+\color{blue}{ 3 \sqrt{7}} \cdot 8 \sqrt{2}+\color{blue}{ 3 \sqrt{7}} \cdot9 = \\ = - 80 \sqrt{2}-90 + 24 \sqrt{14} + 27 \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( 8 \sqrt{2}-9\right) } \cdot \left( 8 \sqrt{2} + 9\right) = \color{blue}{ 8 \sqrt{2}} \cdot 8 \sqrt{2}+\color{blue}{ 8 \sqrt{2}} \cdot9\color{blue}{-9} \cdot 8 \sqrt{2}\color{blue}{-9} \cdot9 = \\ = 128 + 72 \sqrt{2}- 72 \sqrt{2}-81 $$ |
| ⑦ | Simplify numerator and denominator |
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