Помогите пожалуйста, срочно!. Created by MariaB31. algebra-ru

Помогите пожалуйста, срочно!...

MariaB31 @MariaB31

October 2021 1 7 Report

Помогите пожалуйста, срочно!

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NNNLLL54

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$log^2\frac{\pi}{6}=a\; \to \; log\frac{\pi}{6}=\pm \sqrt{a}\\\\\frac{log_2(arccos(-\frac{1}{2}))-3}{log_2\pi -log_23-1}=\frac{log_2(\pi -\frac{\pi}{3})-log_28}{log_2\frac{\pi}{3}-log_22}=\frac{log_2\frac{2\pi }{3\cdot 8}}{log_2\frac{\pi}{3\cdot 2}}=\frac{log_2\frac{\pi}{12}}{log_2\frac{\pi}{6}}=\\\\=\frac{log_2(\frac{1}{2}\cdot \frac{\pi}{6})}{log_2\frac{\pi}{6}}=\frac{log_2\frac{1}{2}+\log_2\frac{\pi}{6}}{log_2\frac{\pi}{6}}=\frac{-1+\log_2\frac{\pi}{6}}{log_2\frac{\pi}{6}}=$

$= \left \{ {{\frac{-1+\sqrt{a}}{\sqrt{a}}} \atop {\frac{-1-\sqrt{a}}{-\sqrt{a}}=\frac{1+\sqrt{a}}{\sqrt{a}}} \right.$

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