`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).`
