(y/y^2+2y-3-y-1/y^2-9): y+1/y^2-4y+3 подробно - вопрос №22312912435542626 от wut123 21.11.2020 10:25

Question

Алгебра: (y/y^2+2y-3-y-1/y^2-9): y+1/y^2-4y+3 подробно распишите

wut123 · Answer

1)y/([(y+3)(y-1)]-(y-1)/[(y-3)(y+3)]=(y²-3y-y²+2y-1)/[(y-1)(y-3)(y+3)]=
=-(y+1)/[(y-1)(y-3)(y+3)]
y²+2y+3=0
y1+y2=-2 U y1*y2=-3⇒y1=-3 U y2=1
2)-(y+1)/[(y-1)(y-3)(y+3)]*(y-1)(y-3)/(y+1)=-1/(y+3)
y²-4y+3=0
y1+y2=4 U y1*y2=3⇒y1=1 U y2=3