Ответы на вопрос:
{10(x+4)(y+1)=30(x-1)(y-1) {30(x-1)(y-1)=20xy 10(x+4)(y+1)=20xy (x+4)(y+1)=2xy xy+x+4y+4=2xy xy-x=4y+4 x(y-1)=(4y+4) x=(4y+4)/(y-1) 10[(4y+4)/(y-1)+4](y+1)=30[(4y+4)/(y-1) -1](y-1) [(4y+4)/(y-1)+4](y+1)=3[(4y+4)/(y-1) -1](y-1) (4y+4+4y-4)(y+1)/(y-1)=3(4y+4-y+1)(y-1)/(y-1) 8y(y+1)=(9y+15)(y-1),y≠1 8y²+8y=9y²-9y+15y-15 y²-2y-15=0 y1+y2=2 u y1*y2=-15 y1=-3⇒x1=(-12+4)/(-3-1)=-8/(-4)=2 y2=5⇒x2=(20+4)/(5-1)=24/4=6 ответ (2; -3); (6; 5)
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