\(f\left(x\right)=x^2-2x+2017\)
\(\Leftrightarrow f\left(x\right)=x^2-x-x+2017\)
\(\Leftrightarrow f\left(x\right)=\left(x^2-x\right)-\left(x-1\right)+2016\)
\(\Leftrightarrow x\left(x-1\right)-\left(x-1\right)+2016\)
\(\Leftrightarrow\left(x-1\right)^2+2016\)
Với mọi x ta có :
\(\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\left(x-1\right)^2+2016\ge2016\)
\(\Leftrightarrow f\left(x\right)\ge0\)
Dấu "=" xảy ra khi :
\(\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)
Vậy ..
ta co : f(x)= x2-2x+2017=x2-2x+1+2016=(x-1)2+2016\(\ge2016\)
dau = xay ra khix=1
Vay ....