$(m-5)x^2-4mx+m-2=0$
$\Delta = b^2-4ac=(-4m)^2-4.(m-5).(m-2)$
$=12m^2+28m-40$
• TH1:
ycbt: $\Leftrightarrow \left\{ \begin{array}{l}\Delta ≥0\\a≠0\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l}12m^2+28m-40 ≥0\\m-5≠0\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l}m ≤-\frac{10}{3} hay 1≤m\\m≠5\end{array} \right.$
• TH2:
$a=0 \Leftrightarrow m-5=0 \Leftrightarrow m=5$
Thay m=5 vào pt:
$\Rightarrow -20x+3=0$
$\Leftrightarrow x=\frac{3}{20}$
$\Rightarrow m=5$ loại
$\Leftrightarrow m ∈ (-∞,-\frac{10}{3}]U[1,+∞)$\{$5$}
