\(x\left(\frac{5-x}{x+1}\right)\left(x+\frac{5-x}{x+1}\right)=6\)
\(x.\frac{5-x}{x+1}.\left(x+\frac{5-x}{x+1}\right)=6\)
\(\Leftrightarrow\frac{x^2\left(5-x\right)}{x+1}+\frac{x\left(5-x\right)^2}{\left(x+1\right)^2}=6\)
\(\Leftrightarrow x^2\left(5-x\right)\left(x+1\right)+x\left(5-x\right)^2=6\left(x+1\right)^2\)
\(\Leftrightarrow5x^3-5x^2-x^4+25x=6x+12x+6\)
\(\Leftrightarrow5x^3-5x^2-x^4+25x-6x^2-12x-6=0\)
\(\Leftrightarrow5x^3-11x^2-x^4+13x-6=0\)
\(\Leftrightarrow\left(x^3-4x^2+7x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+3\right)\left(x-2\right)\left(x-1\right)=0\)
Mà \(x^2-2x+3\ne0\) nên:
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)