1. 4x^3+2x^2+16x+8
2. ax-bx-ay+by+az-bz
thx
數學高手請來
2007-01-23 2:39 am
回答 (4)
2007-01-23 2:47 am
1. 4X^3+2X^2+16X+8
2X^2(2X^2+1)+16X+8
2X^2(2X^2+1)+4(4X+2)
2X^2(2X^2+1)(4X^2)+2
2X^2(2X^2+1+2)+2
2X^2(2X^2+1+2+2)
2.X(A-B)-Y(A-B)+Z(A-B)
X-Y+Z(A-B)
(有好大可能係錯)
(只供參考)
2X^2(2X^2+1)+16X+8
2X^2(2X^2+1)+4(4X+2)
2X^2(2X^2+1)(4X^2)+2
2X^2(2X^2+1+2)+2
2X^2(2X^2+1+2+2)
2.X(A-B)-Y(A-B)+Z(A-B)
X-Y+Z(A-B)
(有好大可能係錯)
(只供參考)
參考: 自己
2007-01-23 2:46 am
1. 4x^3+2x^2+16x+8
= 2(2x^3+x^2+8x+4)
= 2[x^2(2x+1)+4(2x+1)]
= 2(x^2+4)(2x+1)
2. ax-bx-ay+by+az-bz
= x(a-b)-y(a-b)+z(a-b)
= (x-y+z)(a-b)
= 2(2x^3+x^2+8x+4)
= 2[x^2(2x+1)+4(2x+1)]
= 2(x^2+4)(2x+1)
2. ax-bx-ay+by+az-bz
= x(a-b)-y(a-b)+z(a-b)
= (x-y+z)(a-b)
參考: 自己計出來
2007-01-23 2:45 am
4x^3+2x^2+16x+8
=2x^2(2x+1)+8(2x+1)
=(2x^2+8)(2x+1)
=2(x^2+4)(2x+1)
ax-bx-ay+by+az-bz
=x(a-b)-y(a-b)+z(a-b)
=(x-y+z)(a-b)
=2x^2(2x+1)+8(2x+1)
=(2x^2+8)(2x+1)
=2(x^2+4)(2x+1)
ax-bx-ay+by+az-bz
=x(a-b)-y(a-b)+z(a-b)
=(x-y+z)(a-b)
2007-01-23 2:45 am
4x^3+2x^2+16x+8
=2x^2( 2x+1 ) + 8( 2x+1 )
=(2x^2+8)( 2x+1 )
=2(x^2+4)( 2x+1 )
ax-bx-ay+by+az-bz
=(a-b)x-(a-b)y+(a-b)z
=(a-b)(x-y+z)
=2x^2( 2x+1 ) + 8( 2x+1 )
=(2x^2+8)( 2x+1 )
=2(x^2+4)( 2x+1 )
ax-bx-ay+by+az-bz
=(a-b)x-(a-b)y+(a-b)z
=(a-b)(x-y+z)
收錄日期: 2021-04-12 23:26:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070122000051KK02974