Question

Tìm m để :

(m + 2)\(x^2\) - 2(m + 2)x + 3m + 4 < 0, ∀x∈R

Answer 1

+) \(m+2=0\Leftrightarrow m=-2\Rightarrow-2< 0\left(TM\right)\)

+) \(m+2\ne0\Leftrightarrow m\ne-2\)

\(\left(m+2\right)x^2-2\left(m+2\right)x+3m+4< 0,\forall x\in R\)

\(\Leftrightarrow\left\{{}\begin{matrix}m+2< 0\\\Delta'< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< -2\\\left(m+2\right)^2-\left(m+2\right)\left(3m+4\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< -2\\-2m^2-6m-4< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< -2\\m\in\left(-\infty;-2\right)\cup\left(-1;+\infty\right)\end{matrix}\right.\)

\(\Rightarrow m\in\left(-\infty;-2\right)\)

Vậy \(m\in(-\infty;-2]\)