Odpowiedzi

2009-12-08T17:24:05+01:00
A) (2a+b)×c-(2ac-bc+1)=2ac+bc-2ac+bc-1=2bc-1
b) 5(x-z)+2(z-x)=5x-5z+2z-2x=3x-3z
c) 2x²(x-5)-3x(x²-3x+5)=2x³-10x²-3x³+9x²-15=-1x³-1x²-15
d) 6(a²+b+⅓)-6(a²-½)-3(b+⅓)=6a²+6b+2-6a²+3-3b-1=3b+4
2009-12-08T17:40:07+01:00
A) (2a+b)*c-(2ac-bc+1)=(2a+b)*c-2ac+bc-1= 2ac+bc-2ac+bc-1=2bc-1
b) 5(x-z)+2(z-x)=5x-5z+2z-2x=3x-3z
c) 2x²(x-5)-3x(x²-3x+5)=2x³-10x²-3x³+9x²-15x=-x³-x²-15x
d) 6(a²+b+⅓)-6(a²-½)-3(b+⅓)=6a²+6b+2-6a²+3-3b-1=3b+4