Упр.1290 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) ((sin(-30°)-cos(30°))^2/(3cos(-45°)sin(45°)-6tg(30°)ctg(60°));
б) (2tg^2(-?/3)cos(?/3)-3ctg(?/6))/(sin(?/2)cos(?/3)tg(?/4)ctg^2(?/6))?

а)

$$ \frac{(\sin(-30^\circ)-\cos 30^\circ)^2}{3\cos(-45^\circ)\sin 45^\circ-6\tg 30^\circ\ctg 60^\circ} = \frac{\left(-\frac12-\frac{\sqrt3}{2}\right)^2}{3\cdot\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}-6\cdot\frac{\sqrt3}{3}\cdot\frac{\sqrt3}{3}} $$

$$ =\frac{\left(\frac{1+\sqrt3}{2}\right)^2}{\frac32-2} = \frac{\frac{1+2\sqrt3+3}{4}}{-\frac12} = \frac{1+\sqrt3}{2}:\left(-\frac12\right) = -2-\sqrt3 $$

б)

$$ \frac{2\tg^2\left(-\frac{\pi}{3}\right)\cos\frac{\pi}{3}-3\ctg\frac{\pi}{6}} {\sin\frac{\pi}{2}\cos\frac{\pi}{3}\tg\frac{\pi}{4}\ctg^2\frac{\pi}{6}} = \frac{2\cdot(-\sqrt3)^2\cdot\frac12-3\sqrt3} {1\cdot\frac12\cdot1\cdot(\sqrt3)^2} $$

$$ =\frac{3-3\sqrt3}{\frac32} = \frac{6-6\sqrt3}{3} = 2-2\sqrt3 $$

Ответ

а) $$-2-\sqrt3$$; б) $$2-2\sqrt3$$.