Odpowiedź:
zad 1
a)(x+2)(y+1)=xy+x+2y+2
b)(2a+1)(3-b)=6a-2ab+3-b
c)(a-1)(a²-b²)=a³-ab²-a²+b²
d)(2x-1)(y-3)=2xy-6x-y+3
zad 2
a)(x+1)(x+2)=x²+2x+x+2=x²+3x+2
b)(a+4)(a-5)=a²-5a+4a-20=a²-a-20
c)(2t+1)(1-3t)=2t-6t²+1-3t=-6t²-t+1
d)(3ab-7)(3-7ab)=9ab-21a²b²-21+49ab=-21a²b²+58ab-21
e)(-x+y)(5x+6y)=-5x²-6xy+5xy+6y²=-5x²+6y²-xy
f)(x+5y)(7x-y)=7x²-xy+35xy-5y²=7x²-5y²+34xy
g)(2x-y)(x+y)=2x²+2xy-xy-y²=2x²+xy-y²
h)(ab-a)(2ab+6a)=2a²b²+6a²b-2a²b-6a²=2a²b²+4a²b-6a²
Szczegółowe wyjaśnienie:
