几道因式分解啊!11

问题描述:

几道因式分解啊!11
x²-y²-x-y
ab-b²+ac-c²-2bc
bc(b+c)+ca(c-a)-ab(a+b)
a²-ab²-2a+4b
a³-2a-2b+b³
a^4+a²b²+b^4
x^4+324
abc-ab-ac+a-bc+b+c-1
x²-7x+6
(1-x²)(1-y²)-4xy
3x³+7x²-4

x²-y²-x-y
=(x+y)(x-y)-(x+y)
=(x+y)(x-y-1)
ab-b²+ac-c²-2bc
=a(b+c)-(b²+2bc+c²)
=a(b+c)-(b+c)²
=(b+c)(a-b-c)
bc(b+c)+ca(c-a)-ab(a+b)
=b²c+bc²+ac²-a²c-a²b-ab²
=c(b²+bc+ac)-a(ac+ab+b²)
=(c-a)(b²+bc+ac)
a²-ab²-2a+4b
=(a²-2a)-b(a²-4)
=a(a-2)-b(a+2)(a-2)
=(a-2)(a-ab-2b)
a³-2a-2b+b³
=(a+b)(a²-ab+b²)-2(a+b)
=(a+b)(a²-ab+b²-2)
a^4+a²b²+b^4
=a^4+2a²b²+b^4-a²b²
=(a²+b²)²-a²b²
=(a²+b²+ab)(a²+b²-ab)
x^4+324
=x^4+36x²+324-36x²
=(x²+18)²-36x²
=(x²+6x+18)(x²-6x+18)
abc-ab-ac+a-bc+b+c-1
=ab(c-1)-a(c-1)-b(c-1)+(c-1)
=(c-1)(ab-a-b+1)
=(c-1)[a(b-1)-(b-1)]
=(c-1)(b-1)(a-1)
x²-7x+6
=(x-1)(a-6)
(1-x²)(1-y²)-4xy
=1-x²-y²+x²y²-4xy
=(x²y²-2xy+1)-(x²+2xy+y²)
=(xy-1)²-(x+y)²
=(xy+x+y-1)(xy-x-y-1)
3x³+7x²-4
=(3x³+3)+(7x²-7)
=3(x+1)(x²-x+1)+7(x+1)(x-1)
=(x+1)(3x²-3x+3+7x-7)
=(x+1)(3x²+4x-4)
=(x+1)(3x-2)(x+2)