100 . a^2-5a+6/a^2+7a+12*a^2+3a/a^2-4a+4 - вопрос №9058324100256271223244 от drart77 12.02.2020 07:17

Question

Алгебра: 100 . a^2-5a+6/a^2+7a+12*a^2+3a/a^2-4a+4

drart77 · Answer

((a²-5a+6)/((a²+7a+12))*((a²+3a)/(a²-4a+4)).
1. (a²-5a+6)/(a²+7a+12)=(a²-3a-2a+6)/(a²+4a+3a+12)=
=(a*(a-3)-2*(a-3))/(a*(a+4)+3*(a+4))=(a-3)*(a-2)/((a+4)*(a+3)
2.(a²+3a)/(a²-2*a*2+2²)=a*(a+3)/(a-2)²
3. (a-3)*(a-2)*a*(a+3)/((a+4)*(a+3)*(a-2)²)=a*(a-3)/((a-2)*(a+4)).