若2x〃-(4m-1)x+2m〃+1(〃表示平方)是一个完全平方式,求m的值

设2x^2-(4m-1)x+2m^2+1=(ax+b)^2=a^2x^2-2abx+b^2
比较系数
2=a^2
1-4m=-2ab (1)
2m^2+1=b^2
(1)^2 4m-1=2ab ,16m^2-8m+1=4a^2b^2
16m^2-8m+1=4(2)(2m^2+1)
16m^2-8m+1=8(2m^2+1)=16m^2+8
-8m+1=8
8m=-7
m=-7/8

2x^2-(4m-1)x+2m^2+1
=2[x^2-(2m-1/2)x+m^2+1/2]
=2[x-(m-1/4)]^2+m+7/8 是完全平方式
则m+7/8=0
所以m=-7/8