Δ=(2m-1)²-4(m²-m+4)=4m²-2m+1-4m²+4m-16=2m-16 pt có 2 ngo ⇔Δ≥0⇔2m-16≥0⇔m≥8 áp dụng ht vi et tacó:$\left \{ {{x1+x2=2m+1} \atop {x1x2=m^2-m+4}} \right.$ theo bài ra ta có:x1x2-2x1-2x2=5 ⇔x1x2-2(x1+x2)-5=0 ⇔m²-m+4-4m-2-4=0 ⇔m²-5m-2=0 ⇔m1=$\frac{5+$\sqrt{33}$ }{2 }$ m2=$\frac{5-$\sqrt{33}$ }{2}$
