Ответ: 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=0y1+y2=-2 U y1*y2=-3⇒y1=-3 U y2=12)-(y+1)/[(y-1)(y-3)(y+3)]*(y-1)(y-3)/(y+1)=-1/(y+3)y²-4y+3=0y1+y2=4 U y1*y2=3⇒y1=1 U y2=3

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