<span>(tgx+)(2cosx-1)=0
ОДЗ sinx>0⇒x∈[πn+π+πn]
tgx+√3=0
tgx=-√3
x=-π/6+πn∉ОДЗ
2cosx-1=0
cosx=1/2
x=π/3+2πn
x=-π/3+2πn∉ОДЗ</span>
1)x^2 - x - 6
D = 1 + 4 * 6 = 25
x1 = (1 + 5) / 2 = 3
x2 = (1 - 5) / 2 = -2
x^2 - x - 6 = (x - 3)(x + 2)
x^2 - 3x - 10
D = 9 + 40 = 49
x1 = (3 + 7) / 2 = 5
x2 = (3 - 7) / 2 = -2
x^2 - 3x - 10 = (x - 5)(x + 2)
Сокращаем на х + 2( х не равно 2), получаем (х-3)/(х-5)
как то так
x^4-y^4=(x^2-y^2)(y^2+y^2)=2*(x^2+y^2)