ĐKXĐ: \(x\le4\)
\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)+6\left(x+1\right)-\left(4-x\right)\sqrt{4-x}-6\sqrt{4-x}=0\)
\(\Leftrightarrow\left(x+1\right)^3+6\left(x+1\right)-\left(\sqrt{4-x}\right)^3-6\sqrt{4-x}=0\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\\sqrt{4-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^3+6a-b^3-6b=0\)
\(\Rightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+6\left(a-b\right)=0\)
\(\Rightarrow\left(a-b\right)\left[\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}+6\right]=0\)
\(\Rightarrow a=b\)
\(\Rightarrow\sqrt{4-x}=x+1\left(x\ge-1\right)\)
\(\Rightarrow4-x=x^2+2x+1\)
\(\Rightarrow x^2+3x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{21}}{2}\\x=\dfrac{-3-\sqrt{21}}{2}< -1\left(loại\right)\end{matrix}\right.\)
