Как решить (2/2+m - m/m-2 - 4m/4-m^²) : (2/2+m - вопрос №55332442224422244 от refoinda 07.09.2022 11:25

Question

Алгебра: Как решить (2/2+m - m/m-2 - 4m/4-m^²) : (2/2+m + m/2-m + 4m/m^²-4)

refoinda · Answer

2/2+m - m/m-2 - 4m/4-m^²=2/2+m +m/2-m - 4m/4-m^²=
(2(2-m)+m(m+2)-4m)/(4-m^2)=(4-2m+m^2+2m-4m)/ (4-m^2)=
(m^2-4m+4)/ (m^2-4)=(m-2)(m-2)/(2-m)(m+2)=
-(m-2)/(m+2).
(2/2+m + m/2-m + 4m/m^²-4)=(2(2-m)-m(2+m)+4m)/ (m^²-4)=(4-2m-2m-m^2+4m)/(m-2)(m+2)=(4-m^2)/(m^2-4)=-1
-(m-2)/(m+2):(-1)= (m-2)/(m+2)