$ x^2 -2mx +2m -1 = 0$
$ \Delta' = m^2 - (2m -1) = m^2 -2m +1 = (m-1)^2$
Vì $\Delta' = (m-1)^2 \ge 0$ nên PT có nghiệm với mọi $m$
$ x_1 = \dfrac{-b' + \sqrt{\Delta}}{a} = \dfrac{m + |m-1|}{1} = m + |m-1|$
$ x_2 = \dfrac{-b' - \sqrt{\Delta}}{a} = \dfrac{m - |m-1|}{1} = m - |m-1|$
Ta có
$ (x_1^2 - 2mx_1 +3)(x_2^2 - 2mx_2 -2) = 24$
$+)$
$ x_1^2 -2mx_1 +3 = (x_1^2 -2mx + m^2) + 3 - m^2 = ( x_1 - m)^2 +3 - m^2 = ( m + |m-1| - m)^2 +3 - m^2$
$ = |m-1|^2 +3 - m^2 = m^2- 2m +1 + 3 - m^2 = -2m +4$ (*)
$+)$
$ x_2^2 -2mx_2 -2 = (x_2 - m)^2 - m^2 -2 = ( m - |m-1| - m)^2 -m^2 -2 = (-|m-1|)^2 - m^2 -2$
$ = m^2 -2m +1 - m^2 -2= -2m -1$ (**)
Từ (*);(**) suy ra $ (-2m+4)(-2m-1) = 24$
$\to (2m-4)(2m+1) = 24$
$\to 4m^2 -6m -4 -24 = 0 \to 4m^2 -6m -28 = 0$
$ \to 2m^2 -3m +14 = 0 \to ( 2m-7)(m+2) = 0$
$\to$ \(\left[ \begin{array}{l}2m-7=0\\\\m+2=0\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}2m=7\\\\m=-2\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}m = \dfrac{7}{2}\\\\m=-2\end{array} \right.\)
Vậy $ m \in \{ -2 ; \dfrac{7}{2} \}$
