Question

ПОЖАЛУЙСТА, ПОМОГИТЕ!!!!!​

Answer 1

Ответ:

$tga=3\\\\1)\ \ \dfrac{4cosa-2sina}{3sina+cosa}=\dfrac{cosa\, (4-2tga)}{cosa\, (3\, tga+1)}=\dfrac{3-2tga}{3\, tga+1}=\dfrac{3-6}{9+1}=\dfrac{-3}{10}=-0,3\\\\\\2)\ \ \dfrac{5cosa+3sina}{3sina-2cosa}=\dfrac{5+3\, tga}{3\, tga-2}=\dfrac{5+9}{9-2}=\dfrac{14}{7}=2$

$3)\ \ \dfrac{4\, cosa-3sina}{3sin^3a+cos^3a}=\dfrac{cosa\, (4-3\, tga)}{cos^3a\, (3\, tg^3a+1)}=\dfrac{4-3\, tga}{cos^2a\, (3\, tg^3a+1)}=\\\\\\=\dfrac{(1+tg^2a)(4-3\, tga)}{3\, tg^3a+1}=\dfrac{(1+9)(4-9)}{3\cdot 27+1}=\dfrac{10\cdot (-5)}{82}=-\dfrac{50}{82}=-\dfrac{25}{41}$

$4)\ \ \dfrac{sina\cdot cosa-2sin^2a}{3sin^2a+2cos^2a}=\dfrac{cos^2a\, (tga-2\, tg^2a)}{cos^2a(3\, tg^2a+2)}=\dfrac{tga-2\, tg^2a}{3\, tg^2a+2}=\\\\\\=\dfrac{3-18}{18+2}=\dfrac{-15}{20}=-0,75$