a)ta có: f(x)=0
=>x²-8x+2032=0
=>x.x-4x-4x+16+2016=0
=>x(x-4)-4(x-4)+2016=0
=>(x-4).(x-4)+2016=0
=>(x-4)²+2016=0
mà (x-4)²≥0
=>(x-4)²+2016>0
=>f(x) ko có ngiệm
b)ta có:f(x)=x²-8x+2032
mà x²-8x+2032=(x-4)²+2016
=>f(x)=(x-4)²+2016
ta có:(x-4)²≥0
=>(x-4)²+2016≥2016
=>f(x) =2016 khi (x-4)²=0
x-4=0
x=0+4
x=4
vậy Min f(x)=2016 khi x=4