А) 6ab-3b² = 3b(2a-b)
Б) ac+bc-3a-3b = c(a+b)-3(a+b) = (c-3)(a+b)
<span>sinx/2*sin3x/2=1/2
1/2[cos(x/2 - 3x/2) - cos(x/2 + 3x/2)] = 1/2
cos(x) - cos(2x) = 1
применяем формулу: (cos2x = 2</span>cos²x - 1)<span>
2cos</span>²x - cosx = 0
cosx(2cosx - 1) = 0
1) cosx = 0
x₁ = π/2 + πk, k∈Z
2) 2cosx - 1 = 0
cosx = 1/2
x = (+ -)arccos(1/2) + 2πn, n∈Z
x₂ = (+ -)(π/3) + 2πn, n∈Z
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=(3x^2+14x)e^x+(x^3+7x^2)e^x