B_4 = 9, q = 1/3. Найти S_5/
S_n = b_1(1 - q^n) / (1 - q) --- формула n-го члена геометрич. прогрессии.
b_4 = b_1 * q^3 ----> b_1 = b_4 / q^3 = 9 / (1/3) = 9*3 = 27/
S_5 = 27(1 - (1/3)^5 / (1 - 1/3) = 27*(1 - 1/243) /(2/3) = 27* (242/243) *(3/2) =
= 121/3 = 40 1/3
Ответ. 40 1/3
(∛x+∛y)/(∛x²-∛xy+∛y²)-(∛x-∛y)/[(∛x-∛y)(∛x+∛y)]=
=(∛x+∛y)/(∛x²-∛xy+∛y²)- 1/(∛x+∛y)=(∛x²+2∛xy+∛y²-∛x²+∛xy-∛y²)/(x-y) =3∛xy/(x-y)
S4=s1+3d
s8=s1+7d
{-8=s1+3d
-
{32=s1+7d
-40=-4d
d=10
-8=s1+30
s1=-30-8=-38
Ответ:s1=-38,d=10
<span>1-cos6x=tg3x
2sin</span>²3x-sin3x/cos3x=0
cos3x≠0⇒x≠π/6+πk/3,k∈z
2sin²3x*cos3x-sin3x=0
sin3x(2sin3xcos3x-1)=0
sin3x=0⇒x=πk/3,k∈z
sin6x-1=0⇒sin6x=1⇒x=π/12+πk/3,k∈z