b1+b2=140, b1+b1q=140, b1(1+q)=140 (1)
b2+b3=105, b1q+b1q^2=105,b1q(1+q)=105 (2)
разделим (2) на (1) получим q=105/140=3/4
b1(1+3/4)=140, b1=140*4/7=80
b2=80*3/4=60
b3=60*3/4=45
F'(x)=Cos(5x+π/3)*(5x+π/3)'=5cos(5x+π/3)
(x^2-25)/(x^3+4x+25)= (x-5)*(x+5)/(x+5)*(x^2-x+5)= (x-5)/(x^2-x+5)