證明：
x²-2mx-3m²+8m-4=0
要使方程有兩個(gè)實(shí)數(shù)根
△=(-2m)²-4(-3m²+8m-4)
=4m²+12m²-32m+16
=16m²-32m+16
=16(m²-2m+1)
=16(m-1)²≥0
所以原方程永遠(yuǎn)有兩個(gè)實(shí)數(shù)根
所以當(dāng)m﹥2時(shí),原方程有兩個(gè)實(shí)數(shù)根
x²-2mx-3m²+8m-4=0,
分解因式得：[x-(3m-2)][x+(m-2)]=0,
x-(3m-2)=0,x+(m-2)=0,
x1=3m-2,x2=2-m,
由題意
3m-2＜5
2-m＞2
或
3m-2＞2
2-m＜5
解得：m＜0或m＞4/3