`1/{x+1}-{2x^2-7}/{x^3+1}=3/{x^2-x+1}`
ĐKXĐ: `x\ne-1`
`1/{x+1}-{2x^2-7}/{x^3+1}=3/{x^2-x+1}`
`⇔1/{x+1}-{2x^2-7}/{(x+1)(x^2-x+1)}=3/{x^2-x+1}`
`⇔{1(x^2-x+1)}/{(x+1)(x^2-x+1)}-{2x^2-7}/{(x+1)(x^2-x+1)}={3(x+1)}/{(x+1)(x^2-x+1)}`
`⇒1(x^2-x+1)-(2x^2-7)=3(x+1)`
`⇔x^2-x+1-2x^2+7=3(x+1)`
`⇔-x^2-x+8=3x+3`
`⇔-x^2-x-3x=3-8`
`⇔-x^2-4x=-5`
`⇔x^2-4x=5`
`⇔x^2-4x-5=0`
`⇔x^2+5x-x-5=0`
`⇔(x^2+5x)-(x+5)=0`
`⇔x(x+5)-(x+5)=0`
`⇔(x+5)(x-1)=0`
`⇔` \(\left[ \begin{array}{l}x+5=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-5 \text{ (tm ĐKXĐ)}\\x=1\text{ (tm ĐKXĐ)}\end{array} \right.\)
Vậy phương trình có tập nghiệm `S={-5;1}`
