1.若m不等于1.則由b^2-4ac=4m^2+5>0.故方程橫有兩個實根x1、x2.若使x1、x2至少有一根為整數(shù)根,則存在整數(shù)k使得4m^2+5=k^2,即判別式可以開方..即（2m）^2+5=k^2.因為2m必為偶數(shù),則k為奇數(shù).經(jīng)探究只有m=±1,k=3才符合題意
因為m≠1,所以m=-1.
2.令 t = √(5x^2 -1) ≥ 0
則 5x^2 = t^2 +1
原方程即 t^2+1 +x - xt - 2 = 0
t^2 - 1 - xt +x =0
(t+1)(t-1) - x(t-1) =0
(t-1)(t+1-x) =0
t = 1 或 t = x-1
t=1即√(5x^2 -1) = 1
解得 x = ± √10 /5
t = x-1 即 √(5x^2 -1) = x-1 ,
得 2x^2 +x-1 =0
解得 x=0.5 ,x = -1
因為0.5和-1都小于1
此時 t = x-1