(m^2+3m/m^2+3m-4 - m^2-4m/m^2-4m+3) / m/m^2+m-12 - вопрос №48980632323424243212 от naithon00 03.01.2020 10:18

Question

Алгебра: (m^2+3m/m^2+3m-4 - m^2-4m/m^2-4m+3) / m/m^2+m-12

naithon00 · Answer

1)(m²+3m)/(m-1)(m+4)-(m²-4m)/(m-1)(m-3)=
=[m(m+3)(m-3)-m(m-4)(m+4)]/(m-1)(m+4)(m-3)=
=[m(m²-9)-m(m²-16)]/(m-1)(m+4)(m-3)=
=m(m²-9-m²+16)/(m-1)(m+4)(m-3)=7m/(m-1)(m+4)(m-3)
2)7m/(m-1)(m+4)(m-3) : m/(m+4)(m-3)=
=7m/(m-1)(m+4)(m-3)*(m+4)(m-3)/m=7/(m-1)