ĐKXĐ: \(x\ne-1\)
Ta có: \(\dfrac{x\left(3-x\right)}{x+1}\cdot\left(x+\dfrac{3-x}{x+1}\right)=2\)
\(\Leftrightarrow\dfrac{x\left(3-x\right)}{x+1}\cdot\left(\dfrac{x+1+3-x}{x+1}\right)=2\)
\(\Leftrightarrow\dfrac{x\left(3-x\right)}{x+1}\cdot\dfrac{4}{x+1}=2\)
\(\Leftrightarrow\dfrac{4x\left(3-x\right)}{\left(x+1\right)^2}=2\)
\(\Leftrightarrow4x\left(3-x\right)=2\left(x+1\right)^2\)
\(\Leftrightarrow12x-4x^2=2\left(x^2+2x+1\right)\)
\(\Leftrightarrow-4x^2+12x=2x^2+4x+2\)
\(\Leftrightarrow-4x^2+12x-2x^2-4x-2=0\)
\(\Leftrightarrow-6x^2+8x-2=0\)
\(\Leftrightarrow-6x^2+6x+2x-2=0\)
\(\Leftrightarrow-6x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-6x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-6x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-6x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{1}{3}\right\}\)
\(\dfrac{x\left(x-3\right)}{x+1}\left(x+\dfrac{3-x}{x+1}\right)=2\)
\(\Leftrightarrow\)\(\left(\dfrac{3x-x^2}{x+1}\right)\left(\dfrac{x\left(x+1\right)}{x+1}+\dfrac{3-x}{x+1}\right)=2\)
\(\Leftrightarrow\)\(\left(\dfrac{3x-x^2}{x+1}\right)\left(\dfrac{x^2+3}{x+1}\right)=2\)
\(\Leftrightarrow\)\(\dfrac{\left(3x-x^2\right)\left(x^2+3\right)}{\left(x+1\right)^2}=2\)
\(\Leftrightarrow\)\(\dfrac{3x^3+9x-x^4-3x^2}{x^2+2x+1}=2\)
\(\Rightarrow3x^3+9x-x^4-3x^2=2x^2+4x+2\)
\(\Leftrightarrow3x^3+9x-x^4-3x^2-2x^2-4x-2=0\)
\(\Leftrightarrow3x^3-x^4+5x-5x^2-2=0\)
\(\Leftrightarrow-\left(x-1\right)^2\left(x^2-x+2\right)=0\)
Vì x^2 -x+2 >0
\(\Rightarrow\left(x-1\right)^2=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy phương trình có nghiệm duy nhất là x=1
