a)\(\frac{x^2+x+1}{x-1}=\frac{x^2-2x+1}{x-1}+\frac{3x}{x-1}=\left(x-1\right)+\frac{3x}{x-1}\)
Để \(x^2+x+1⋮x-1\) thì \(\frac{3x}{x-1}\) nguyên.Tức là
\(x-1\inƯ\left(3x\right)\Leftrightarrow1-\frac{1}{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Giải tiếp:v