Odpowiedź :
[tex](a+b)(c+d) = ac + ad + bc + bd[/tex]
[tex](a+b)(c+d+e) = ac+ad+ae+bc+bd+be[/tex]
1.
[tex]a) \ (x+2)(y+1) = xy + x + 2y + 2\\\\b) \ (2a+1)(3-b) = 6a-2ab + 3 -b[/tex]
2.
[tex]a) \ (x+1)(x+2) = x^{2}+2x+x+2 =x^{2}+3x+2\\\\b) \ (3a+4)(a-5) = 3a^{2}-15a + 4a - 20 = 3a^{2}-11a - 20\\\\c) \ (2t + 1)(1-3t) = 2t -6t^{2}+1-3t = -6t^{2}-t+1\\\\d) \ (3ab-7)(3-7ab) =9ab - 21a^{2}b^{2}-21+49ab = -21a^{2}b^{2}+58ab-21[/tex]
3.
[tex]a) \ (t + 1)(t^{2}+2t+3) = t^{3}+2t^{2}+3t + t^{2}+2t+3 = t^{3}+3t^{2}+5t+3\\\\b) \ (p-2)(3p^{2}-5p+1) = 3p^{3}-5p^{2}+p - 6p^{2}+10p-2 = 3p^{3}-11p^{2}+11p-2[/tex]
Odpowiedź:
Cześć
Szczegółowe wyjaśnienie:
Zad 1
a) (x+2)(y+1)=xy+x+2y+2
b)(2a+1)(3-b)=6a-2ab+3-b
Zad 2
a) x²+2x+x+2
b) 3a²-15a+4a-20=3a²-11a+4
c)2t-6t²+1-3t=-1t-6t²+1
d) 9ab-21a²b²-21-49ab