Question

сократите дробь $frac{3a^4+a^3+6a^2+a+3}{3a^3-2a^2+2a-3}$

Answer 1

(3a^4+a^3+6a^2+a+3)/(3a^3-2a^2+2a-3)=(3a^4+a^3+3a^2+3a^2+a+3)/((3(a^3-1)-2a(a-1))=(3a^2(a^2+1)+a(a^2)+1+3(a^2+1))/(3(a-1)(a^2+a+1)-2a(a-1))=((a^2+1)(3a^2+a+3))/((a-1)(3a^2+3a+3-2a)=(a^2+1)(a-1)