Ax+ay/x+y+xy=
=a(x+y)/x+y(1+x)=
a/(1+a)
1) 12a^3-16a^2=4a^2*(3a-4)
2) 5m^2+15m^3-20m^5=5m^2*(1+3m-4m^3)
3) Вы пропустили степень у переменной Х во втором слагаемом
Решение
2sinxcos3x + sin4x = 0
2*(1/2)*[sin(x - 3x) + sin(x + 3x)] + sin4x = 0
- sin2x + 2sin4x = 0
2sin2x*cos2x - sin2x = 0
sin2x*(2cos2x - 1) = 0
1) sin2x = 0
2x = πk, k ∈ Z
x₁ = πk/2, k ∈ Z
2) 2cos2x - 1 = 0
cos2x = 1/2
2x = (+ -)*arccos(1/2) + 2πn, n ∈ Z
2x = (+ -)*(π/3) + 2πn, n ∈ Z
x₂ = (+ -)*(π/6 + πn, n ∈ Z
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