Đáp án:
Tìm x
`x^3-16x=0`
`=> x(x^2-16)=0`
`=> x(x-4)(x+4)=0`
`=>` \(\left[ \begin{array}{l}x=0\\x+4=0\\x-4=0\end{array} \right.\) 
`=>` \(\left[ \begin{array}{l}x=0\\x=-4\\x=4\end{array} \right.\) 
`A/(x-1)^3+B/(x-1)^2+C/(x-1)`
`<=>A/(x-1)^3+(B.(x-1))/(x-1)^3+(C.(x-1)^2)/(x-1)^3`
`<=>(A+Bx-B+Cx2-2Cx+1)/(x-1)^3`
ta có: `A+Bx-B+Cx^2-2Cx+1=x^2-x+2`
`<=>` `Cx^2=x^2 => C=1`
      `Bx-2Cx=-x =>Bx-2x=-x => Bx=x => B=1`
      `A+1-B=2 => A+1-1=2 => A=2`
Vậy: `A=2; B=1 và C=1`