Đáp án+Giải thích các bước giải:
ĐKXĐ:$\left\{\begin{matrix}1-x\ne0\\(x-1).(x^2+x+1)\ne0\\x^2+x+1\ne0\end{matrix}\right.$
`<=>`$\left\{\begin{matrix}1-x\ne0\\x-1\ne0\\x^2+x+1\ne0\end{matrix}\right.$
`<=>`$\left\{\begin{matrix}1-x\ne0\\x-1\ne0\\(x+\dfrac{1}{2})^2+\dfrac{3}{4}\ne0(Lđ)\end{matrix}\right.$
`<=>x\ne1`
Với `x\ne1` ta có:
`1+(x-2)/(1-x)+(2x^2-5)/(x^3-1)=(4)/(x^2+x+1)`
`<=>(x^3-1)/(x^3-1)-(x-2)/(x-1)+(2x^2-5)/(x^3-1)=(4)/(x^2+x+1)`
`<=>(x^3-1-(x-2).(x^2+x+1)+2x^2-5)/(x^3-1)=(4.(x-1))/(x^3-1)`
`<=>x^3-1-(x^3+x^2+x-2x^2-2x-2)+2x^2-5=4x-4`
`<=>x^3-1-x^3-x^2-x+2x^2+2x+2+2x^2-5-4x+4=0`
`<=>3x^2-3x=0`
`<=>3.x.(x-1)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0(tm)\\x=1(Loại)\end{array} \right.\)
Vậy `S={0}`
