x^4+x^3+2x^2+x+1-1-2x-x^2+y^2x-查字典问答网
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来自陈建党的问题

  x^4+x^3+2x^2+x+1-1-2x-x^2+y^2x^2+ax^2+x+ax-1-ax^3-2x^2-x+2+x^5-2x^4x^3+x^2-y^3-y^2ab(c^2-d^2)-(a^2-d^2)cdx^8-y^82(x+y)+6(x+y)^2-4(x+3)^3-(3a^2-5b^2)^2+(5a^2-3b^2)^2(5x^2+2x-3)^2-(x^2-2x-3)^2x^2(a+b)^2-2xy(a^2-b^2)+y(a-b)^29a^2+x^2n+6a+2x^n+6ax^n+1

  x^4+x^3+2x^2+x+1

  -1-2x-x^2+y^2

  x^2+ax^2+x+ax-1-a

  x^3-2x^2-x+2+x^5-2x^4

  x^3+x^2-y^3-y^2

  ab(c^2-d^2)-(a^2-d^2)cd

  x^8-y^8

  2(x+y)+6(x+y)^2-4(x+3)^3

  -(3a^2-5b^2)^2+(5a^2-3b^2)^2

  (5x^2+2x-3)^2-(x^2-2x-3)^2

  x^2(a+b)^2-2xy(a^2-b^2)+y(a-b)^2

  9a^2+x^2n+6a+2x^n+6ax^n+1

1回答
2020-07-31 13:17
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孙龙杰

  x^4+x^3+2x^2+x+1

  =x^4+x^3+x^2+x^2+x+1

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

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

  -1-2x-x^2+y^2

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

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

  =(y+x+1)(y-x-1)

  x^2+ax^2+x+ax-1-a

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

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

  x^3-2x^2-x+2+x^5-2x^4

  =x^5-2x^4+x^3-2x^2-x+2

  =x^4(x-2)+x^2(x-2)-(x-2)

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

  x^3+x^2-y^3-y^2

  =x^3-y^3+x^2-y^2

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

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

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

  ab(c^2-d^2)-cd(a^2-b^2)

  =abc^2-abd^2-a^2cd+b^2cd

  =(abc^2-a^2cd)-abd^2+b^2cd

  =ac(bc-ad)+bd(bc-ad)

  =(bc-ad)(ac+bd)

  x^8-y^8

  =(x^4+y^4)(x^4-y^4)

  =(x^4+y^4)(x^2+y^2)(x^2-y^2)

  =(x^4+y^4)(x^2+y^2)(x+y)(x-y)

  2(x+y)+6(x+y)^2-4(x+3)^3

  =2(x+y)[1+3(x+y)-4(x+y)^2]

  =2(x+y)[1-4(x+y)][1+(x+y)]

  =2(x+y)(1-4x-4y)(1+x+y)

  -(3a^2-5b^2)^2+(5a^2-3b^2)^2

  =(5a^2-3b^2)^2-(3a^2-5b^2)^2

  =[(5a^2-3b^2)-(3a^2-5b^2)][(5a^2-3b^2)+(3a^2-5b^2)]

  =(2a^2+2b^2)(8a^2-8b^2)

  =16(a^2+b^2)(a+b)(a-b)

  (5x^2+2x-3)^2-(x^2-2x-3)^2

  =(5x^2+2x-3)^2-(x^2-2x-3)^2

  =[(5x^2+2x-3)-(x^2-2x-3)][(5x^2+2x-3)+(x^2-2x-3)]

  =(6x^2+4x)(6x^2-6)

  =12x(3x+2)(x+1)(x-1)

  x^2(a+b)^2-2xy(a^2-b^2)+y^2(a-b)^2

  =[x(a+b)-y(a-b)]^2

  9a^2+x^2n+6a+2x^n+6ax^n+1

  =x^2n+6ax^n+9a^2+6a+2x^n+1

  =(x^n+3a)^2+2(x^n+3)+1

  =(x^n+3a+1)^2

2020-07-31 13:19:22

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