Question

将下列各式分解因式 :a^2-b^2-4a+4b ac+2ab-bc-a^2-b^2 a^2+b^2-c^2-d^2-2ab+2cd x^2+y^2-1-2xy

还有(2x^2-3x)(2x^2-3x-1)-2
a^2-b^2-4a+4b ac+2ab-bc-a^2-b^2
a^2+b^2-c^2-d^2-2ab+2cd x^2+y^2-1-2xy

Answer 1

通过配方法来解决，(ax)^2+2abx+b^2=（ax+b）^2;(ax)^2-2abx+b^2=(ax-b)^2看上面的式子，配凑出这样的形式，缺少的项加上，就多加了，减去同样的式子就行了，依次类推。

Answer 2

a^2-b^2-4a+4b
=a^2-4a+4-b^2+4b-4
=(a+2)^2-(b-2)^2
=[(a+2)+(b-2)][(a+2)-(b-2)]
=(a+b)(a-b+4)

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

a^2+b^2-c^2-d^2-2ab+2cd
=a^2-2ab+b^2-c^2+2cd-d^2
=(a-b)^2-(c-d)^2
=[(a-b)+(c-d)][(a-b)-(c-d)]
=(a-b+c-d)(a-b-c+d)

x^2+y^2-1-2xy
=x^2-2xy+y^2-1
=(x-y)^2-1
=(x-y+1)(x-y-1)

Answer 3

a^2-b^2-4a+4b=（a+2)^2-(b-2)^2=(a+b)(a-b+4)
ac+2ab-bc-a^2-b^2=c（a-b）-(a^2+b^2-2ab)=c(a-b)-(a-b)^2=(a-b)(c-a+b)
a^2+b^2-c^2-d^2-2ab+2cd=（a-b)^2-(c-d)^2=(a-b+c-d)(a-b-c+d)
x^2+y^2-1-2xy=(x-y)^2-1=(x-y+1)(x-y-1)

Answer 4

(2x^2-3x)(2x^2-3x-1)-2=(2x^2-3x)^2-(2x^2-3x)-2
=(2x^2-3x-2)(2x^2-3x+1)
=(2x+1)(x-2)(2x-1)(x-1)