Решение:
1) f'(x)=-1/(1-x)=1/(x-1)
k=-1 1/(x-1)=-1 x-1=-1 x=0
f(0)=ln1=0
y=-x
2) k=3
3/(3x-2)=3 3=9x-6 9=9x x=1
f(1)=ln1=0
y=3x+c 0=3*1+c c=-3
y=3x-3
3) f'=(2x-2)/(x^2-2x-3)
3y=1-2x y=-2/3x+1/3
(x-1)/(x^2-2x-3)=-1/3
3x-3=-x^2+2x+3
x^2+x-6=0
x=-3 x=2
y(2)=ln(-3) не существует
y(-3)=ln(9-3+6)=ln12
ln12=-2/3(-3)+c
ln12=2+c x=ln12-2
y=-2/3x+ln12-2
4) (-2x-2)/(3-2x-x^2)
2/3x-1/3=y
(x+1)/(x^2+2x-3)=1/3
3x+3=x^2+2x+3
x^2-x=0
x=1 x=0
f(0)=ln3
y=2/3x+c
y=2/3x+ln3.
4000⁻³ = 1/4000³ = 1/(4³*10⁹) = 1/64 * 10⁻⁹ = 1,5625*10⁻¹¹