Question

[Hà Như Thuỷ]

Xác định m để tam thức f(x) = x² + (3 - m)x - 2m + 3 luôn luôn dương với mọi \(x\le-4\)

Answer 1

[Nguyễn Việt Lâm]

\(a=1>0\) ; \(\Delta=\left(3-m\right)^2-4\left(-2m+3\right)=m^2+2m-3\)

Để \(f\left(x\right)>0\) ; \(\forall x\le-4\)

TH1: \(\Delta< 0\Leftrightarrow m^2+2m-3< 0\Leftrightarrow-3< m< 1\)

TH2: \(\left\{{}\begin{matrix}\Delta=0\\-\frac{b}{2a}>-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m^2+2m-3=0\\\frac{m-3}{2}>-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=1\\m=-3\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}\Delta>0\\-4< x_1< x_2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2+2m-3>0\\\left(x_1+4\right)\left(x_2+4\right)>0\\\frac{x_1+x_2}{2}>-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2+2m-3>0\\x_1x_2+4\left(x_1+x_2\right)+16>0\\x_1+x_2>-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2+2m-3>0\\-2m+3+4\left(3-m\right)+16>0\\m-3>-8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2+2m-3>0\\-6m+31>0\\m>-5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>1\\m< -3\end{matrix}\right.\\m< \frac{31}{6}\\m>-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-5< m< -3\\1< m< \frac{31}{6}\end{matrix}\right.\)

Kết hợp lại ta được: \(-5< m< \frac{31}{6}\)