Помогите пожалуйста решить очень подробно

Для начала лучше упростить
x≠4
f(x)= (x^2(x-4)+x-4) / (x-4) = (x-4)(x^2+1) = x^2+1
f(1)=1+1 = 2
f'(x)=2x
f'(-2)=-2*2=-4
-4+2=-2
D)

Формулы:

(u / v)' = (u'*v - u*v') / v^2

(u + v)' = u' + v'

f(x) = (x^3 - 4x^2 + x - 4) / (x - 4) =
= [x^2 * (x - 4) + (x - 4)] / (x - 4) =
= x^2 + 1

f(x) = (x^3 - 4x^2 + x - 4) / (x - 4) =

f'(x) = [(x^3 - 4x^2 + x - 4)' *(x - 4) - (x^3 - 4x^2 + x - 4)*(x - 4)'] / (x - 4)^2 =
= [(3x^2 - 8x + 1) * (x - 4) - (x^3 - 4x^2 + x - 4) * 1] / (x - 4)^2 =
= [(3x^2 - 8x + 1)/(x - 4)] - [(x^3 - 4x^2 + x - 4)/(x - 4)^2] =
f(-2) = [(3*(-2)^2 - 8*(-2) + 1)/((-2) - 4)] -
- [((-2)^3 - 4(-2)^2 + (-2) - 4)/((-2) - 4)^2] =
= [(12 + 16 + 1)/(-6)] - [(-8) - 16 - 2 - 4)/36] =
= 29/(-6) - (-30)/36 = - 29/6 + 5/6 = - 24/6 = - 4


f(x) = (x^3 - 4x^2 + x - 4) / (x - 4) =
f(1) = (1^3 - 4*1^2 + 1 - 4) / (1 - 4) =
= (1 - 8 - 3) / (-3) = 10/3

f(-2) + f(1) = - 4 + 10/3 = - 2/3