ĐKXĐ: x∉-1
Ta có: \(x\cdot\frac{3-x}{x+1}\cdot\left(x+\frac{3-x}{x+1}\right)=2\)
\(\Leftrightarrow\frac{x\left(3-x\right)}{x+1}\cdot\left(\frac{x\left(x+1\right)}{x+1}+\frac{3-x}{x+1}\right)=2\)
\(\Leftrightarrow\frac{x\left(3-x\right)}{x+1}\cdot\frac{x^2+3}{x+1}-2=0\)
\(\Leftrightarrow\frac{x\left(3-x\right)\cdot\left(x^2+3\right)}{\left(x+1\right)^2}-\frac{2\left(x+1\right)^2}{\left(x+1\right)^2}=0\)
\(\Leftrightarrow\left(3x-x^2\right)\left(3+x^2\right)-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow9x+3x^3-3x^2-x^4-2x^2-4x-2=0\)
\(\Leftrightarrow-x^4+3x^3-5x^2+5x-2=0\)
\(\Leftrightarrow-\left(x^4-3x^3+5x^2-5x+2\right)=0\)
\(\Leftrightarrow x^4-x^3-2x^3+5x^2-5x+2=0\)
\(\Leftrightarrow x^3\left(x-1\right)+5x\left(x-1\right)-\left(2x^3-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+5x\right)-2\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+5x\right)-2\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x^3+5x\right)-2\left(x^2+x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+5x-2x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2+3x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-x+2\right)=0\)(1)
Ta có: \(x^2-x+2=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\forall x\)(2)
Từ (1) và (2) suy ra \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)(tm)
Vậy: x=1