4.а)2х²-(4-2х)=2х²+(2х-4)
в)4х+(-2х+1)=4х-(2х-1)
5.а)...+(-2у+3з)
б)...-(2у-3з)
.......................................................................
B3=b1*q^2
1=4*q^2
0.25=q^2
q=0.5
Решение
cos(πx/6 + 5/6π) = - 1/2
πx/6 + 5/6π = (+ -) arccos(- 1/2) + 2πk, k ∈ Z
πx/6 + (5/6)π = (+ -) (2π/3) + 2<span>πk, k ∈ Z
</span>πx/6 = (+ -) (2π/3) - (5/6)π + 2πk, k ∈ Z
πx = (+ -) (4π) - 5π + 12<span>πk, k ∈ Z
</span>x = (+ -) (4) - 5π + 12<span>k, k ∈ Z</span>