Odpowiedź :
Odpowiedź:
a)(a+3b)(7a+2b)=7a²+2ab+21ab+6b²=7a²+23ab+6b²
b)5-(x+3)(x+1)=5-(x²+x+3x+3)=5-x²-4x-3=-x²-4x+2
c)t-(t+3)(t-5)=t-(t²-5t+3t-15)=t-t²+5t-3t+15=-t²+3t+15
Szczegółowe wyjaśnienie:
[tex]a) \ (a+3b)(7a+2b) = a\cdot7a + a\cdot2b + 3b\cdot7a + 3b\cdot2b = 7a^{2}+2ab + 21ab +6b^{2} =\\\\= 7a^{2}+23ab + 6b^{2}[/tex]
[tex]b) \ 5 - (x+3)(x+1) = 5 - (x\cdot x + x\cdot1 + 3\csot x + 3\cdot1) = 5 - (x^{2}+x+3x+3) =\\\\=5-(x^{2}+4x+3) = 5 - x^{2}-4x-3 = -x^{2}-4x + 2[/tex]
[tex]c) \ t - (t+3)(t-5) = t - [t\cdot t +t\cdot(-5)+3\cdot t + 3\cdot(-5)] = t - (t^{2}-5t+3t-15)=\\\\=t - (t^{2}-2t-15) = t-t^{2}+2t+15 = -t^{2}+3t+15[/tex]