1-cos2x=sin2x
2sin²x=sin2x
2sin²x-2sinxcosx=0
sinx(sinx-cosx)=0
sinx=0
x=πk
sinx-cosx=0
tgx=1
x=π/4+πk
Ответ: x=πk, x=π/4+πk; k∈Z
( 3x + 3y) + (bx+by) = 3(x+y)+ b(x+y) = (3+ b) (x+y)
X³(x - 2) + 2x²(x - 2) - 21x(x - 2) + 18(x - 2) = 0
(x - 2)*(x³ + 2x² - 21x + 18) = 0
(x - 2) *(x²(x - 1) + 3x(x - 1) - 18(x - 1)) = 0
(x - 2)*(x - 1)*(x² + 3x - 18) = 0
(x - 2)*(x - 1)*(x(x - 3) + 6(x - 3)) = 0
(x - 2)*(x - 1)*(x - 3)*(x + 6) = 0
x - 2 = 0
x = 2
x - 1 = 0
x = 1
x - 3 = 0
x = 3
<span>x + 6 = 0
x = - 6</span>
3x/(x-4)-6x/(x-4)^2=(3x(x-4)-6x)/(x-4)^2=(3x^2-18x)/(x-4)^2=3x(x-6)/(x-4)^2