$x^{2}$ -mx + 2×(m-2)=0
Δ=($-m^{2}$) -4.1.2(2m-2)
= $m^{2}$ -8m +16
= $m-4^{2}$ $\geq$ 0 ∀ m
Ta có
x1 + x2 = m
x1x2= 2m-4
$\left \{ {{2x1 + 3x2= 5} \atop {x1 + x2 = m}} \right.$
$\left \{ {{2(x1+x2)+x2=5} \atop {x1+x2=m}} \right.$
$\left \{ {{x2=5-2m} \atop {x1=3m-5}} \right.$
x1x2=(3m-5)(5-2m)
⇔2m-4=15m-25-$6m^{2}$ +10m ⇔ 6$m^{2}$ -23m +21=0
m1=$\frac{7}{3}$
m2=$\frac{3}{2}$