初一数学因式分解问题 求一些初一的数学因式分解难题（可附答案，只要好就采纳！）不要...
\u521d\u4e00\u6570\u5b66\u56e0\u5f0f\u5206\u89e3\u95ee\u9898\u4e0b\u5217\u5404\u5f0f\u4ece\u5de6\u5230\u53f3\u7684\u53d8\u5f62\u5c5e\u4e8e\u56e0\u5f0f\u5206\u89e3\u7684\u662f\uff1f\uff08c\uff09
a
(x+3)(x-3)=x^2-9
bx^2-2x+1=x(x-2)+1
c
x(x-4y)+4y^2=(x-2y)^2
dx^3+5x-24=(x+3)(x-8)
\u628a\u591a\u9879\u5f0f(1+x)(1-x)-(x-1)\u63d0\u516c\u56e0\u5f0f(x-1)\u540e\uff0c\u4f59\u4e0b\u7684\u90e8\u5206\u662f\uff1f\uff08d\uff09
a(x+1)
b-(x+1)
cx
d-(x+2)
10^11-5x10^9(\u7528\u56e0\u5f0f\u5206\u89e3\u8981\u8fc7\u7a0b\uff09
=10^9(10^2-5)
=10^9*(100-5)
=95*10^9
=9.5*10^10
lz\u597d\uff0c\u6211\u8fd9\u513f\u6709\u4e00\u4e9b\u9898\u76ee\uff1a
1\u3001a^2+a-b^2-b=(a+b)(a-b)+(a-b)
=(a-b)(a+b+1)
2\u3001a^2*b^2-a^2-b^2+1=a^2(b^2-1^2)-(b^2-1^2)
=(a^2-1^2)(b^2-1^2)
=(a+1)(a-1)(b+1)(b-1)
3\u3001(ax+by)^2+(ay-bx)^2=(ax)^2+2axby+(by)^2+(ay)^2-2aybx+(bx)^2
=(ax)^2+(ay)^2+(by)^2+(bx)^2
=a^2(x^2+y^2)+b^2(y^2+x^2)
=(a^2+b^2)(x^2+y^2)
4\u3001a^2-2ab+b^2-c^2-2c-1=(a-b)^2-(c+1)^2
=(a-b+c+1)(a-b-c-1)
5\u3001x^4+x^2*y^2+y^4=(x^2+y^2)-(xy)^2
=(x^2+y^2+xy)(x^2+y^2-xy)
6\u30014x^2-4x-y^2+4y-3=(4x^2-4x+1)-(y^2-4y+4)
=(2x-1)^2-(y-2)^2
=(2x+y-3)(2x-y+1)
7\u3001x^4+2x^3+3x^2+2x+1=x^4+x^3+x^2
x^3+x^2+x
x^2+x+1
=x^2(x^2+x+1)+x(x^2+x+1)+(x^2+x+1)
=(x^2+x+1)^2
8\u3001a^3+b^3+c^3-3abc=a^3+3a^2*b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc
=(a+b)^3+c^3-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2+2ab-ac-bc)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
\u624b\u6253\u4e0d\u5bb9\u6613\uff0c\u671b\u91c7\u7eb3\uff0c\u5efa\u8baelz\u591a\u4e70\u53c2\u8003\u4e66\uff0c\u4e0d\u8fc71-2\u672c\u5c31\u591f\u4e86
1、9x^2m-3x^m
=3x^m(x^2-1)
=3x^m(3x^m-1)
2、∵a+b=2,ab=-1/2
∴a(a+b)(a-b)-a(a+b)^2
=a(a+b)(a-b-a-b)
=a(a+b)(-2b)
=-2ab(a+b)
=-2*(-1/2)*2
=2
3、∵a^2+a+1=0
∴a^3+2a^2+2a+1
=a^3+a^2+a+a^2+a+1
=a(a^2+a+1)+a^2+a+1
=(a^2+a+1)(a+1)
=0*(a+1)
=0
∴a^2001+a^2002+...+a^2009
=a^2001(1+a+a^2+...+a^8)=a^2001[(1+a+a^2)+(a^3+a^4+a^5)+(a^6+a^7+a^8)]
=a^2001(1+a+a^2)(1+a^3+a^6)
＝0
1.A=a+c,B=b
2.-(3x+y)
3.35
4.(x-y)²(a²-b²)
5.(5a²-2b²+2a²-5b²)(5a²-2b²-(2a²-5b²))=(7a²-7b²)(3a²+3b²)=21(a²-b²)(a²+b²)
6.0.01*(67²-33²)=0.01*(67+33)(67-33)=34
7.(4
2/3+45
1/3)(4
2/3-45
1/3)=50*(-41
1/3)=-2066
1/3
8.5a²-20b²=5(a²-4b²)=5(a+2b)(a-2b)=5*5*3=75
9.因为a+b=5，a²-b²=5，所以a-b=1,再与a+b=5构成方程组，解得a=3,b=2
10.由(N+1998)²=65432178，得出N+1998，
(N+1998)(N+2008)=(N+1998)(N+1998+100)=(N+1998)²+100(N+1998)，代入即可
11.
①原式=-（x-y）²
②题目不对啊
③原式=-（m-2n）²
④
⑤
高毅
这么简单也不会啊
对了
你的诛仙0.0
（x-1）（x-3）-15
=x^2-4x+3-15
=x^2-4x-12
=（x-6)(x+2)
先把前面的展开，就变成x的平方-4x-12=
(x-6)(x+2)
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