\(\left\{{}\begin{matrix}-1\le\dfrac{m-2}{m+1}\\\dfrac{m-2}{m+1}\le1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}0\le1+\dfrac{m-2}{m+1}\\\dfrac{m-2}{m+1}-1\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}0\le\dfrac{2m-1}{m+1}\\\dfrac{-3}{m+1}\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}0\le\dfrac{2m-1}{m+1}\\m+1>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}0\le2m-1\\m+1>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\ge\dfrac{1}{2}\\m>-1\end{matrix}\right.\)\(\Rightarrow m\ge\dfrac{1}{2}\)

\(-1\le\dfrac{m-2}{m+1}\le1\)

⇔ \(\left|\dfrac{m-2}{m+1}\right|\le1\)

⇔ \(\dfrac{\left(m-2\right)^2}{\left(m+1\right)^2}\le1\)

⇔ \(\left\{{}\begin{matrix}m^2-4m+4\le m^2+2m+1\\m\ne-1\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}m\ge\dfrac{1}{2}\\m\ne-1\end{matrix}\right.\)⇔ m ≥ \(\dfrac{1}{2}\)

Tập nghiệm: \(S=[\dfrac{1}{2};+\infty)\)

m 2   -   ( 2 m   -   1 ) ( m   +   1 )   <   0

⇔ - m 2   -   m   +   1   <   0

( 2 m   -   1 ) 2   -   4 ( m   +   1 ) ( m   -   2 )   ≥   0  ⇔ 9 ≥ 0. Bất phương trình có tập nghiệm là R.

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