初二10道因式分解的数学题,要过程

如题所述

第1个回答  2012-02-14
(1) 6x4y2-12x3y+27x2y3

=3x2y·2x2y-3x2y·4x+3x2y·9y2=3x2y(2x2y-4x+9y2)

(2)-x4y+x3y2-x2y3=-(x4y-x3y2+x2y3)

=-(x2y·x2-x2y·xy+x2y·y2)

=-x2y(x2-xy+y2)

(3)xn+3xn-1+xn-2=xn-2·x2+xn-2·3x+xn-2·1

=xn-2(x2+3x+1)

(4)5(x-y)3+10(y-x)2=5(x-y)3+10(x-y)2

=5(x-y)2(x-y+2)

(5)m(5ax+ay-1)-m(3ax-ay-1)

=m[(5ax+ay-1)-(3ax-ay-1)]

=m·(5ax+ay-1-3ax+ay+1)

=m(2ax+2ay)=2ma(x+y)
(6)x2+14x+49=x2+2·x·7+72

=(x+7)2

(7)(m+n)2-6(m+n)+9=(m+n-3)2

(8)-3ax2+6axy-3ay2=-3a(x2-2xy+y2)

=-3a(x-y)2

(9)(x2+4)2+8x(x2+4)+16x2

=(x2+4)2+2·(x2+4)·4x+(4x)2

=(x2+4+4x)2

=(x+2)4

(10)64m2n2-(m2+16n2)2

=(8mn)2-(m2+16n2)2

=(8mn+m2+16n2)(8mn-m2-16n2)

=(m+4n)2[-(m-4n)2]

=-(m+4n)2(m-4n)2本回答被网友采纳
第2个回答  2012-02-15
(3)xn+3xn-1+xn-2=xn-2·x2+xn-2·3x+xn-2·1

=xn-2(x2+3x+1)
第3个回答  2012-02-14
哎 题目呢

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