Bài 1: Tìm x (3x-1)^2-(3x-1)(3x+3)(3x+3)=8 Bài 2: tính 4(x-y)^3-(x-y)(x^2+xy+y^2)

1 bình luận về “Bài 1: Tìm x (3x-1)^2-(3x-1)(3x+3)(3x+3)=8 Bài 2: tính 4(x-y)^3-(x-y)(x^2+xy+y^2)”

1. Zuiiyy

   Bài 1:

   (3x-1)^2-(3x-1)(3x+3)(3x+3)=8

   <=>(3x)^2-2.3x+1-(3x-1)(3x+3)(3x+3)=8

   <=>3^2.x^2-2.3x+1-(3x-1)(3x+3)(3x+3)=8

   <=>9x^2-2.3x+1-(3x-1)(3x+3)(3x+3)=8

   <=>9x^2-6x+1-(3x-1)(3x+3)(3x+3)=8

   <=>9x^2-6x+1-(9x^2-3x+9x-3)(3x+3)=8

   <=>9x^2-6x+1-27x^3+9x^2-27x^2+9x-27x^2+9x-27x+9=8

   <=>-36x^2-15x+10-27x^3=8

   <=>-36x^2-15x+10-27x^3-8=0

   <=>-36x^2-15x+2-27x^3=0

   <=>x=0,1048

   Vậy x=0,1048

                                                                                  

   Bài 2:

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

   =4x^3-12x^2 y+12xy^2-4y^3-(x-y)(x^2+xy+y^2)

   =4x^3-12x^2 y+12xy^2-4y^3-x^3+y^3

   =y^3-4y^3+4x^3-x^3-12x^2 y+12xy^2

   =-3y^2+4x^3-x^3-12x^2 y+12xy^2

   =-3y^3+3x^3-12x^2 y+12xy^2

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