ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\x\ne-1\end{matrix}\right.\)
Ta có: \(M=\frac{x^2-2x}{x^3+1}+\frac{1}{2}\left(\frac{1}{1+\sqrt{x+2}}+\frac{1}{1-\sqrt{x+2}}\right)\)
\(=\frac{x^2-2x}{x^3+1}+\frac{1}{2}\cdot\left(\frac{1-\sqrt{x+2}+1+\sqrt{x+2}}{1-\left(x+2\right)}\right)\)
\(=\frac{x^2-2x}{x^3+1}+\frac{1}{2}\cdot\left(\frac{2}{1-x-2}\right)\)
\(=\frac{x^2-2x}{x^3+1}+\frac{-1}{x+1}\)
\(=\frac{x^2-2x}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^2-2x-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{-\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=-\frac{1}{x^2-x+1}\)
Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow x^2-x+1\ge\frac{3}{4}\forall x\)
\(\Rightarrow\frac{1}{x^2-x+1}\le\frac{4}{3}\forall x\)
\(\Rightarrow-\frac{1}{x^2-x+1}\ge-\frac{4}{3}\forall x\)
Dấu '=' xảy ra khi \(x-\frac{1}{2}=0\)
hay \(x=\frac{1}{2}\)
Vậy: Giá trị nhỏ nhất của biểu thức \(M=\frac{x^2-2x}{x^3+1}+\frac{1}{2}\left(\frac{1}{1+\sqrt{x+2}}+\frac{1}{1-\sqrt{x+2}}\right)\) là \(-\frac{4}{3}\) khi \(x=\frac{1}{2}\)
