需要初一数学因式分解公式法题目20道及答案，谢谢

需要初一数学因式分解公式法题目20道及答案，谢谢

1.因式分解3a^3b^2c－6a^2b^2c^2＋9ab^2c^3＝3ab^2 c(a^2-2ac+3c^2)
2.因式分解xy＋6－2x－3y＝(x-3)(y-2)
3.因式分解x^2(x－y)＋y^2(y－x)＝(x+y)(x-y)^2
4.因式分解2x^2－(a－2b)x－ab＝(2x-a)(x+b)
5.因式分解a^4－9a^2b^2＝a^2(a+3b)(a-3b)
8.因式分解ab(x^2－y^2)＋xy(a^2－b^2)＝(ay+bx)(ax-by)
9.因式分解(x＋y)(a－b－c)＋(x－y)(b＋c－a)＝2y(a-b-c)
10.因式分解a^2－a－b^2－b＝(a+b)(a-b-1)
11.因式分解(3a－b)2－4(3a－b)(a＋3b)＋4(a＋3b)2＝[3a-b-2(a+3b)]^2=(a-7b)^2
12.因式分解(a＋3)2－6(a＋3)＝(a+3)(a-3)
(13)3x^2－6x＝3x(x-2)
(14)49x^2－25＝(7x+5)(7x-5)
(15)6x^2－13x＋5＝(2x-1)(3x-5)
(16)x^2＋2－3x＝(x-1)(x-2)
(17)12x^2－23x－24＝(3x-8)(4x+3)
(18)(x＋6)(x－6)－(x－6)＝(x-6)(x+5)
(19)3(x＋2)(x－5)－(x＋2)(x－3)＝2(x-6)(x+2)
(20)9x^2＋42x＋49＝(3x+7)^2追问可以补充下过程吗追答不懂十字相乘法？？我帮不了你，你可以从简单的完全平方，平方差入手，再多练十字相乘法的，比如x∧2-x-6=（x＋3）（x-2）多看，寻找熟悉感1. 5ax+5bx+3ay+3by
解法：=5x(a+b)+3y(a+b)
=(5x+3y)(a+b)
2. x^3-x^2+x-1
解法：=(x^3-x^2)+(x-1)
=x^2(x-1)+ (x-1)
=(x-1)(x^2+1)
3. x2-x-y2-y
解法：=(x2-y2)-(x+y)
=(x+y)(x-y)-(x+y)
=(x+y)(x-y-1)

bc(b+c)+ca(c-a)-ab(a+b)
=bc(c-a+a+b)+ca(c-a)-ab(a+b)
=bc(c-a)+bc(a+b)+ca(c-a)-ab(a+b)
=bc(c-a)+ca(c-a)+bc(a+b)-ab(a+b)
=(bc+ca)(c-a)+(bc-ab)(a+b)
=c(c-a)(b+a)+b(a+b)(c-a)
=(c+b)(c-a)(a+b)．

x^2+3x-40
=x^2+3x+2.25-42.25
=(x+1.5)^2-(6.5)^2
=(x+8)(x-5)．

(x^2+x+1)(x^2+x+2)-12时，可以令y=x^2+x,则
原式=(y+1)(y+2)-12
=y^2+3y+2-12=y^2+3y-10
=(y+5)(y-2)
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)．

(1+y)^2-2x^2(1+y^2)+x^4(1-y)^2

解：原式=(1+y)^2+2(1+y)x^2(1+y)+x^4(1-y)^2-2(1+y)x^2(1-y)-2x^2(1+y^2)

=[(1+y)+x^2(1-y)]^2-2(1+y)x^2(1-y)-2x^2(1+y^2)

=[(1+y)+x^2(1-y)]^2-(2x)^2

=[(1+y)+x^2(1-y)+2x]·[(1+y)+x^2(1-y)-2x]

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

=[(x+1)^2-y(x^2-1)][(x-1)^2-y(x^2-1)]

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

x^5+3x^4y-5x^3y^2+4xy^4+12y^5

解：原式=(x^5+3x^4y)-(5x^3y^2+15x^2y^3)+(4xy^4+12y^5)

=x^4(x+3y)-5x^2y^2(x+3y)+4y^4(x+3y)

=(x+3y)(x^4-5x^2y^2+4y^4)

=(x+3y)(x^2-4y^2)(x^2-y^2)

=(x+3y)(x+y)(x-y)(x+2y)(x-2y)

分解因式m +5n-mn-5m
解：m +5n-mn-5m= m -5m -mn+5n
= (m -5m )+(-mn+5n)
=m(m-5)-n(m-5)
=(m-5)(m-n)

分解因式bc(b+c)+ca(c-a)-ab(a+b)
解：bc(b+c)+ca(c-a)-ab(a+b)=bc(c-a+a+b)+ca(c-a)-ab(a+b)
=bc(c-a)+ca(c-a)+bc(a+b)-ab(a+b)
=c(c-a)(b+a)+b(a+b)(c-a)
=(c+b)(c-a)(a+b)