PTDTTNT: (x+3)^2(3x+8)(3x+10)-8 (2x-1)(x-1)(x-3)(2x+3)+9 (4x+1)(12x-1)(3x+2)(x+1)-4

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1. `1)(x+3)^2(3x+8)(3x+10)-8`

   `=(x^2+6x+9)(9x^2+54x+80)-8`

   Đặt `y=x^2+6x+9⇒9y=9.(x^2+6x+9)=9x^2+54x+81`

   `⇒ 9y-1=9x^2+54x+81-1`

   `⇔9y-1= 9x^2+54x+80`

   `⇒bt⇔y.(9y-1)-8`

   `=9y^2-y-8`

   `=9y^2-9y+8y-8`

   `=9y(y-1)+8(y-1)`

   `=(y-1)(9y+8).`

   Thay `y=x^2+6x+9` vào lại biểu thức ta được: 

   `bt⇔(x^2+6x+9-1).[9.(x^2+6x+9)+8]`

   `=(x^2+6x+8).(9x^2+54x+81+8)`

   `=(x^2+6x+8).(9x^2+54x+89)`

   `=(x^2+2x+4x+8).(9x^2+54x+89)`

   `=[x(x+2)+4(x+2)].(9x^2+54x+89)`

   `=(x+2)(x+4)(9x^2+54x+89).`

   `2) (2x-1)(x-1)(x-3)(2x+3)+9`

   `=[(2x-1)(x-1)][(x-3)(2x+3)]+9`

   `=(2x^2−3x+1)(2x^2−3x-9)+9`

   Đặt `2x^2-3x-4=y`

   `⇒bt⇔(y+5)(y-5)+9`

   `=y^2-25+9`

   `=y^2-16`

   `=(y-4)(y+4)`

   Thay `2x^2-3x-4=y` lại vào biểu thức ta có:

   `bt⇔(2x^2-3x-4-4)(2x^2-3x-4+4)`

   `=(2x^2-3x-8)(2x^2-3x)`

   `=x(2x^2-3x-8)(2x-3).`

   `3)(4x+1)(12x-1)(3x+2)(x+1)-4`

   `=[(4x+1)(3x+2)][(12x-1)(x+1)]-4`

   `=(12x^2+11x+2)(12x^2+11x−1)-4`

   Đặt `y=12x^2+11x+2`

   `⇒bt⇔y(y-3)-4`

   `=y^2-3y-4`

   `=y^2+y-4y-4`

   `=y(y+1)-4(y+1)`

   `=(y+1)(y-4)`

   Thay `y=12x^2+11x+2` vào lại biểu thức ta có:

   `bt⇔(12x^2+11x+2+1)(12x^2+11x+2-4)`

   `=(12x^2+11x+3)(12x^2+11x-2).`

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