ĐKXĐ: \(-1\le x\le4\)
\(\Leftrightarrow3x^3-6x^2-9x+\left(6-x-3\sqrt{4-x}\right)+x+3-3\sqrt{x+1}=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(3x+3\right)+\frac{x^2-3x}{6-x+3\sqrt{4-x}}+\frac{x^2-3x}{x+3+3\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(3x+3+\frac{1}{6-x+3\sqrt{4-x}}+\frac{1}{x+3+3\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x^2-3x=0\)