Odpowiedź :
zad 1
3(1+×)(2-×)+4(×-1)(×+2)=3(2-x+2x-x2)+4(x2+2x-x-2)=6-3x+6x-3x2+4x2+8x-4x-8= x2+7x-2
Zad 2
-5(2y+1)(y-6)-(3y-4)(5+y)=-5(2y2-12y+y-6)-(15y+3y2-20-4y)=-10y2+60y-5y+30+15y-3y2+20+4y=-7y2+74y+50
zad 1
3(1+×)(2-×)+4(×-1)(×+2)=3(2-x+2x-x2)+4(x2+2x-x-2)=6-3x+6x-3x2+4x2+8x-4x-8= x2+7x-2
Zad 2
-5(2y+1)(y-6)-(3y-4)(5+y)=-5(2y2-12y+y-6)-(15y+3y2-20-4y)=-10y2+60y-5y+30+15y-3y2+20+4y=-7y2+74y+50