Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
m{x^2} + \left( {m - 1} \right)x + m - 1 < 0,\forall x \in R\\
TH1:m = 0 \Rightarrow 0{x^2} - x - 1 < 0 \Leftrightarrow - x - 1 < 0 \Leftrightarrow x > - 1\left( {KTM} \right)\\
TH2:m \ne 0\\
m{x^2} + \left( {m - 1} \right)x + m - 1 < 0,\forall x \in R\\
\Leftrightarrow dths\,y = m{x^2} + \left( {m - 1} \right)x + m - 1\,nam\,hoan\,toan\,phia\,duoi\,truc\,hoanh\\
\Leftrightarrow \left\{ \begin{array}{l}
m < 0\\
\Delta = {\left( {m - 1} \right)^2} - 4m\left( {m - 1} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < 0\\
\left( {m - 1} \right)\left( {m - 1 - 4m} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < 0\\
\left( {m - 1} \right)\left( { - 3m - 1} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < 0\\
m > 1;m < - \frac{1}{3}
\end{array} \right. \Leftrightarrow m < - \frac{1}{3}
\end{array}$
