Bài 1:
`A=125x^3-10x^2+2x-1`
`=125x^3+15x^2-25x^2+5x-3x-1`
`=(125x^3+15x^2+5x)-(25x^2+3x+1)`
`=5x(25x^2+3x+1)-(25x^2+3x+1)`
`=(5x-1)(25x^2+3x+1)`
`B=x^4+4`
`=x^4+4x^2-4x^2+4`
`=(x^4+4x^2+4)-4x^2`
`=[(x^2)^2+2.x^2 .2+2^2]-4x^2`
`=(x^2+2)^2-(2x)^2`
`=(x^2+2x+2)(x^2-2x+2)`
`C=x^4+4y^4`
`=x^4+4y^4+4x^2y^2-4x^2y^2`
`=(x^4+4x^2y^2+4y^4)-4x^2y^2`
`=[(x^2)^2+2.x^2 .2y^2+(2y^2)^2]-(2xy)^2`
`=(x^2+2y^2)^2-(2xy)^2`
`=(x^2+2xy+2y^2)(x^2-2xy+2y^2)`
Bài 2:
`a)x^3-7x^2-9x+63=0`
`⇔(x^3-7x^2)-(9x-63)=0`
`⇔x^2(x-7)-9(x-7)=0`
`⇔(x-7)(x^2-9)=0`
`⇔(x-7)(x^2-3^2)=0`
`⇔(x-7)(x+3)(x-3)=0`
`⇔`$\left[\begin{matrix} x-7=0\\ x+3=0\\x-3=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=7\\ x=-3\\x=3\end{matrix}\right.$
Vậy `x∈{7;-3;3}`
