Упр.1296 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) v(3/4+2cos^2(30°))+v(5/4-3tg^2(30°));
б) v(tg^2(?/3)-2 3/4)-2/3v(1/2sin^2(?/4)+2)?

а)

$$ \sqrt{\frac{3}{4}+2\cos^2 30^\circ}+\sqrt{\frac{5}{4}-3\tg^2 30^\circ} $$
$$ \cos 30^\circ=\frac{\sqrt{3}}{2}, \qquad \tg 30^\circ=\frac{\sqrt{3}}{3} $$
$$ \sqrt{\frac{3}{4}+2\left(\frac{\sqrt{3}}{2}\right)^2}+\sqrt{\frac{5}{4}-3\left(\frac{\sqrt{3}}{3}\right)^2} = \sqrt{\frac{3}{4}+\frac{3}{2}}+\sqrt{\frac{5}{4}-1} $$
$$ = \sqrt{\frac{9}{4}}+\sqrt{\frac{1}{4}} = \frac{3}{2}+\frac{1}{2} =2 $$

б)

$$ \sqrt{\tg^2 \frac{\pi}{3}-2\frac{3}{4}}-\frac{2}{3}\sqrt{\frac{1}{2}\sin^2 \frac{\pi}{4}+2} $$
$$ \tg \frac{\pi}{3}=\sqrt{3}, \qquad \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2} $$
$$ \sqrt{(\sqrt{3})^2-\frac{11}{4}}-\frac{2}{3}\sqrt{\frac{1}{2}\left(\frac{\sqrt{2}}{2}\right)^2+2} = \sqrt{3-\frac{11}{4}}-\frac{2}{3}\sqrt{\frac{1}{4}+2} $$
$$ = \sqrt{\frac{1}{4}}-\frac{2}{3}\sqrt{\frac{9}{4}} = \frac{1}{2}-\frac{2}{3}\cdot\frac{3}{2} = \frac{1}{2}-1 = -\frac{1}{2} $$

Ответ

а) $$2$$; б) $$-\frac{1}{2}$$.