1. Берілген өрнекті дәреже түрінде жазып, мәнін табыңдар: 1) 125(5a 3b3)-2 а-2b4, мұндағы а= 0,2, b = 0,5;
2) (0,5а 2) -2 : (32a*b*), мұндағы а = (0,5) 4, b = 0,25;
3) (23a 3b)-1. 64а 4 : аг, мұндағы а= -0,125, b = 0,5;
4) 27 (–32a3): (35a1b2)3, мұндағы а= -0,1, b = 0,1.​

Решение
1)  2cosx-1 < 0
cosx < 1/2
arccos(1/2) + 2πn < x < 2π - arccos(1/2) + 2πn, n ∈ Z
π/3 + 2πn < x < 2π - π/3 + 2πn, n ∈ Z
π/3 + 2πn < x < 5π/3 + 2πn, n ∈ Z
2)  sin2x - √2/2 < 0
 sin2x < √2/2 
- π - arcsin(√2/2) + 2πk < 2x < arcsin(√2/2) + 2πk, k ∈ Z
- π - π/4 + 2πk < 2x < π/4 + 2πk, k ∈ Z
 - 5π/4 + 2πk < 2x < π/4 + 2πk, k ∈ Z
 - 5π/8 + πk < x < π/8 + πk, k ∈ Z
3)  tgx<1
- π/2 + πn < x < arctg(1) + πn, n ∈ Z
- π/2 + πn < x < π/4 + πn, n ∈ Z

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