Х/x+2-x/x-2=1+2-1-2=<span>0
</span>1/x^2-9+2x-3/x-3=<span>1/x^2+2x-3/x-12=(1+2x^3-3x-12x^2)/x^2</span>
Решение
<span>F(x)=2x^3-4x^2-5x+3, f'(2)-?
f`(x) = 6x</span>² - 8x - 5
f`(2) = 6*2² - 8*2 - 5 = 24 - 16 - 5 = 3
Решение задания приложено
(1/3*1/2m+1/3*(-3)-m*1/2m-m*(-3)
(1/3*1/2m+1*(-1)-m*1/2m-m*(-3)
(m/2*3+1*(-1)-m*1/2m-m*(-3)
1/6m+1*(-a)-m=1/2m-m*(-3)
(1/6m-1-m*1/2m-m*(-3)
(1/6m-1-1/2m^2-m*(-3)
(1/6m-1-1/2m^2+3m)
(-1/2m^2+1/6m+3m-1)
(-1/2m^2+1/6m+3m*6/6-1)
(-1/2m^2+1/6m+18m/6-1)
(-1/2m^2+1/6(m+18m)-1)
(-1/2m^2+19m/6-1)
-1/2m^2+19m/6-1 ответ!