`x^2+x+3/(x^2+x+1)=3`
`<=>x^2(x^2+x+1)+x(x^2+x+1)+3=3(x^2+x+1)`
`<=>x^4+2x^3+2x^2+x+3=3x^2+3x+3`
`<=>x^4+2x^3−x^2−2x=0`
`<=> x(x^3+2x^2-x-2)=0`
TH1: `x=0`
TH2: `x^3+2x^2-x-2=0`
`<=>x^2(x+2)-(x+2)=0`
`<=>(x+2)(x-1)(x+1)=0`
`<=>` \(\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy `S={-2;-1;0;1}`.
\(x^2+x+\dfrac{3}{x^2+x+1}=3\)
\(\Leftrightarrow x^2+x+1+\dfrac{3}{x^2+x+1}=4\)
\(\Leftrightarrow\left(x^2+x+1\right)^2-4\left(x^2+x+1\right)+3=0\)
\(\Leftrightarrow\left(x^2+x+1-3\right)\left(x^2+x+1-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\x\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=0\\x=-1\end{matrix}\right.\).
Vậy...
\(x^2+x+\dfrac{3}{x^2+x+1}=3\)
\(\Leftrightarrow\left(x^2+x+1\right)+\dfrac{3}{x^2+x+1}=4\)
\(\Leftrightarrow\left(x^2+x+1\right)^2+3-4\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[\left(x^2+x+1\right)-3\right]\left[\left(x^2+x+1\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=3\\x^2+x+1=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x=0\end{matrix}\right.\)\(\Rightarrow x\in\left\{1;-2;0;-1\right\}\)
VẬy pt có tập nghiệm \(S=\left\{1;-2;0;-1\right\}\)
