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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 - 10x - 3x³ + 9x² -15x =
= -x³ +11x² - 25x
d) 6(a²+b+⅓)-6(a²-½)-3(b+⅓) =
= 6a² + 6b + 2 - 6a² + 3 -3b - 1 =
= 3b +1
b) 5(x-z)+2(z-x) = 5x - 5z + 2z - 2x = -3z + 3x
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 + 3 - 6a² + 3 - 3b - 1,5 = 3b + 4,5