Đáp án:
$\begin{array}{l}
1){x^2} + x - 6 < 0\\
\Leftrightarrow {x^2} + 3x - 2x - 6 < 0\\
\Leftrightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right) < 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {x - 2} \right) < 0\\
+ TH1:\left\{ \begin{array}{l}
x + 3 < 0\\
x - 2 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x < - 3\\
x > 2
\end{array} \right.\left( {ktm} \right)\\
+ TH2:\left\{ \begin{array}{l}
x + 3 > 0\\
x - 2 < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > - 3\\
x < 2
\end{array} \right.\\
\Leftrightarrow - 3 < x < 2\\
Vậy\, - 3 < x < 2\\
2)\left( {m - 2} \right).x = m + 1\left( 1 \right)\\
+ Khi:m - 2 = 0 \Leftrightarrow m = 2\\
\left( 1 \right):0.x = 2 + 1 = 3\left( {ktm} \right)\\
+ Khi:m\# 2\\
x = \dfrac{{m + 1}}{{m - 2}} > 0\\
+ TH1:\left\{ \begin{array}{l}
m + 1 > 0\\
m - 2 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m > - 1\\
m > 2
\end{array} \right. \Leftrightarrow m > 2\\
+ TH2:\left\{ \begin{array}{l}
m + 1 < 0\\
m - 2 < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
m < 2
\end{array} \right. \Leftrightarrow m < - 1\\
Vậy\,m < - 1\,hoac\,m > 2
\end{array}$
