Question

因式分解求解答

Answer 1

分解因式a³+b³+c³-3abc
解：a³+b³+c³-3abc=a²(a+b+c)-a²b-a²c+b²(a+b+c)-b²a-b²c+c²(a+b+c)-ac²-bc²-3abc
=(a+b+c)(a²+b²+c²)-(a²b+b²a+abc)-(abc+b²c+bc²)-(a²c+abc+ac²)
=(a+b+c)(a²+b²+c²)-ab(a+b+c)-bc(a+b+c)-ac(a+b+c)
=(a+b+c)(a²+b²+c²-ab-bc-ac)
=(1/2)(a+b+c)[(a-b)²+(b-c)²+(c-a)²]

Answer 2

a^3+b^3+c^3-3abc
=(a^3+3a^2b+3ab^2+b^3+c^3)-(3abc+3a^2b+3ab^2)
=[(a+b)^3+c^3]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2+2ab-3ab-ac-bc)
=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)

Answer 3

(1) x - x + 4x -4 = 0 x （x-1)+4(x-1)=0 (x-1)(x +4)=0 x=1 (2) 6x - 13.5 = 0 x =9/4 x=±1.5 (3) 2x

Answer 4

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