ĐKXĐ: \(-2\le x\le8\)
\(\Leftrightarrow\sqrt{x+2}-3+1-\sqrt{8-x}=3x^3-21x^2+2x-14\)
\(\Leftrightarrow\frac{x-7}{\sqrt{x+2}+3}+\frac{x-7}{1+\sqrt{8-x}}=\left(x-7\right)\left(3x^2+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\frac{1}{\sqrt{x+2}+3}+\frac{1}{1+\sqrt{8-x}}=3x^2+2\left(1\right)\end{matrix}\right.\)
Xét (1), do \(\left\{{}\begin{matrix}\sqrt{x+2}\ge0\\\sqrt{8-x}\ge0\end{matrix}\right.\) \(\Rightarrow VT< \frac{1}{3}+1< 2\)
\(VP=3x^2+2\ge2>VT\)
\(\Rightarrow\) (1) vô nghiệm
Vậy pt có nghiệm duy nhất \(x=7\)
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