初二10道因式分解的数学题,要过程
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(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
=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
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(3)xn+3xn-1+xn-2=xn-2·x2+xn-2·3x+xn-2·1
=xn-2(x2+3x+1)
=xn-2(x2+3x+1)
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哎 题目呢
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