Mathematics: How do you solve this?

I think it would be option (D)....

How do you solve this? - id15261305, meowmeowcow

Ответ:

mookiejacks

07.02.2022

$(\sqrt{3}-\sqrt{6}+\sqrt{12}-\sqrt{24})\cfrac{\sqrt{6}}{2}\qquad \begin{cases} 12=2\cdot 2\cdot 3\\ \qquad 2^2\cdot 3\\ 24=2\cdot 2\cdot\cdot 2 3\\ \qquad 2^2\cdot 6 \end{cases}$

$(\sqrt{3}-\sqrt{6}+\sqrt{2^2\cdot 3}-\sqrt{2^2\cdot 6})\cfrac{\sqrt{6}}{2}\implies (\sqrt{3}-\sqrt{6}+2\sqrt{3}-2\sqrt{6})\cfrac{\sqrt{6}}{2} \\\\\\ (3\sqrt{3}-3\sqrt{6})\cfrac{\sqrt{6}}{2}\implies \cfrac{(3\sqrt{3}-3\sqrt{6})\sqrt{6}}{2}\implies \cfrac{3\sqrt{18}-3\sqrt{6^2} }{2} \\\\\\ \cfrac{3\sqrt{3^2\cdot 2}-3\cdot 6}{2}\implies \cfrac{3\cdot 3\sqrt{2}-18}{2}\implies \cfrac{9\sqrt{2}-18}{2}\implies \cfrac{9\sqrt{2}}{2}-9$

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How do you solve this? $(\sqrt{3} - \sqrt{6} + \sqrt{12} -\sqrt{24})*\frac{\sqrt{6}}{2}$

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