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b) 5(x-z)+2(z-x)
c) 2x²(x-5)-3x(x²-3x+5)
d) 6(a²+b+⅓)-6(a²-½)-3(b+⅓)
a) (2a+b)*c-(2ac+bc-1)= 2ac + bc - 2ac -bc +1 = 1
b) 5x-5z+2z-2x = 3x-3z
c) 2x³-10x²-3x³+9x²-15x = -x³-x²-15x
d) 6a²+6b+2 - 6a²+3 - 3b-1 = 3b-4
akurat to przerabiamy więc chyba jest dobrze
pozdrawiam ;**