ĐK: \(x\neq -1\)
pt \(\Leftrightarrow\dfrac{x^2\left(3-x\right)}{x+1}+\dfrac{x\left(3-x\right)^2}{\left(x+1\right)^2}-2=0\)
\(\Leftrightarrow\dfrac{x^2\left(3-x\right)\left(x+1\right)+x\left(3-x\right)^2-2\left(x+1\right)^2}{\left(x+1\right)^2}=0\\ \Leftrightarrow x^2\left(-x^2+2x+3\right)+x\left(9-6x+x^2\right)-2\left(x^2+2x+1\right)=0\\ \Leftrightarrow-x^4+2x^3+3x^2+9x-6x^2+x^3-2x^2-4x-2=0\\ \Leftrightarrow x^4-3x^3+5x^2-5x+2=0\\ \Leftrightarrow\left(x^4-2x^3+x^2\right)-\left(x^3-2x^2+x\right)+\left(2x^2-4x+2\right)=0\\ \Leftrightarrow x^2\left(x^2-2x+1\right)-x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x^2-x+2\right)=0\)
Mà \(x^2-x+2=\left(x-x+\dfrac{1}{4}\right)+\dfrac{7}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\) \(\forall\) \(x\)
\(\Rightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) (TM)
Vậy \(S=\left\{1\right\}\)
