A1) (x^3-10x^2+5x+1)' = 3x^2-20x+5
f'(1) = 3-20+5
f'(1) = -12
A2) (4/x^4+2K3)' = -16/x^5
f'(2) = -16/32
f'(2) = -0.5
A3) f'(x) = ((x^5-3)Kx)' = 5x^4 * Kx + (x^5-3)/2Kx
A4) f'(x) = ((5-x)/(x+2))' = (-x+2-5+x)/(x+2)^2 = -3/(x+2)^2
B1) (-2/x-1/x^2-3)' = 0
(2x+2)/x^3 = 0
2x+2 = 0
x = -1
x^3 не= 0
х не=0, следовательно х = -1