问题:

急等!因式分解(2-3y)(9y^2加6y加4)与(x加2)^3-(x-2)^3与2分之1x^4-8与x^4+7x^2-8还有a^4-a^3加a^2-a与4a^2-9b^2加c^2-4ac与(ax加by)^2加(bx-ay)^2我给20分快

更新时间:2024-04-24 12:04:58

问题描述:

急等!因式分解

(2-3y)(9y^2加6y加4)与(x加2)^3-(x-2)^3与2分之1x^4-8与x^4+7x^2-8

还有a^4-a^3加a^2-a与4a^2-9b^2加c^2-4ac与(ax加by)^2加(bx-ay)^2我给20分快

刘行华回答:

  (2-3y)(9y^2+6y+4)=(2-3y)(3y+2)^2

  (x+2)^3-(x-2)^3

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

  =4(3x^2+4)

  1/2*x^4-8

  =1/2*(x^4-16)

  =1/2*(x^2-4)(x^2+4)

  =1/2*(x-2)(x+2)(x^2+4)

  x^4+7x^2-8

  =(x^2-1)(x^2+8)

  =(x-1)(x+1)(x^2+8)

  a^4-a^3+a^2-a

  =a^3(a-1)+a(a-1)

  =a(a-1)(a^2+1)

  4a^2-9b^2+c^2-4ac

  =4a^2-4ac+c^2-(3b)^2

  =(2a-c)^2-(3b)^2

  =(2a-c+3b)(2a-c-3b)

  (ax+by)^2+(bx-ay)^2

  =(ax)^2+2abxy+(by)^2+(bx)^2-2abxy+(ay)^2

  =(ax)^2+(by)^2+(bx)^2+(ay)^2

  =a^2(x^2+y^2)+b^2(x^2+y^2)

  =(a^2+b^2)(x^2+y^2)

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