Решение 1. Область определения y = 2cos(x-п/3) D(y) = R 2. Область значения - 1 ≤ 2cos(x-п/3) ≤ 1 - 1/2 ≤ cos(x-п/3) ≤ 1/2 1) cos(x-п/3) ≥ - 1/2 - arccos(-1/2) + 2πk ≤ x - п/3 ≤ arccos(-1/2) + 2πk, k ∈ Z - 2π/3 + 2πk ≤ x - п/3 ≤ 2π/3 + 2πk, k ∈ Z - 2π/3 + π/3 + 2πk ≤ x ≤ 2π/3 + π/3 + 2πk, k ∈ Z - π/3 + 2πk ≤ x ≤ π + 2πk, k ∈ Z 2) cos(x-п/3) ≤ - 1/2 arccos(-1/2) + 2πk ≤ x - п/3 ≤ 2π - arccos(-1/2) + 2πk, k ∈ Z 2π/3 + 2πk ≤ x - п/3 ≤ 2π - 2π/3 + 2πk, k ∈ Z 2π/3 + 2πk ≤ x - п/3 ≤ 4π/3 + 2πk, k ∈ Z 2π/3 + π/3 + 2πk ≤ x ≤ 4π/3 + π/3 + 2πk, k ∈ Z π + 2πk ≤ x ≤ 5π/3 + 2πk, k ∈ Z
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1) -7а(а³+6а²-а+2)=-7а⁴-42а³+7а²-14а
2) (5+х)(х-3)=5х-15+х²-3х=х²+2х-15
3)(6-5а)(9b+3)=54b+18-45ab-15а
4)8a(a+1)-(a+1)=8а²+8-а-1=8а²-а+7
5)2a(1-b)-3(b-1)=2а-2аb-3b+3
6) 9x-17=2x+4 9х-2х=4+17 7х=21 х=3