ХРИСТА РАДИ Ученик по алгебре за 3 четверть получил оценки...​

36a^4 - 25 = (6a^2)^2 - 5^2 = (6a^2 - 5)(6a^2 + 5)
216x^3 - 1 = (6x)^3 - 1^3 = (6x-1)(36x^2+6x+1)
100b^2 - 140bx^2 + 49x^4 = (10b - 7x^2)^2=(10b-7x^2)(10b-7x^2)
125b^3 + 27 = (5b + 3)(25b^2 - 15b + 9)
(5a - 1/5)^2 = 25a^2 - 2a + 1/25)
(3a - 5b^2)(9a^2 + 15ab^2 + 25b^4) = (3a)^3 - (5b^2)^3 = 27a^3 - 125b^6
(0,8x+ 5)(5 - 0,8x) = (5 + 0,8x)(5 - 0,8x) = 5^2 - (0,8x)^2  = 25 - 0,64x^2
(7x+ 0,4)^2 = 49x^2 + 5,6x + 0,16
(6y + 1)(36y^2 - 6y + 1) = (6y)^3 + 1^3 = 216y^3 + 1
25x^2 + 60xy + 36y^2 = (5x + 6y)^2 = (5x + 6y)(5x + 6y).

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