a是一元二次方程x²+3x+1=0的根

∵b=√(2-a)+√(a-2)-3
∴﹛2-a≥0
a-2≥0
∴a=2
此时b=-3
代入方程得 2x²-3x+c=0
x=1代入得 2-3+c=0
c=1
∴方程是2x²-3x+1=0
1.∵a是方程x²+3x+1=0的根
∴a²+3a+1=0
a²+3a=-1
[(a²-4)/(a²-4a+4)-1/(2-a)]÷2/(a²-2a)
=[(a-2)(a+2)/(a-2)²+1/(a-2)]×(a²-2a)/2
=(a+3)/(a-2)×a(a-2)/2
=a(a+3)/2
=(a²+3a)/2
=-1/2

2）方程有：a^2+3a+1=0, 即a^2+3a=-1
原式=[(a+2)(a-2)/(a-2)^2-1/(2-a)]÷2/[a(a-2)]
=[(a+2)/(a-2)+1/(a-2)]÷2/[a(a-2)]
=(a+3)÷(2/a)
=a(a+3)/2
=(a^2+3a)/2
=-1/2

3)为使b有意义，根号下须为非负数