Ta\(\Leftrightarrow x=\dfrac{1}{2}\) có:
\(P=2017-\left(x-\dfrac{1}{2}\right)^2=-\left(x-\dfrac{1}{2}\right)^2+2017\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+2017\le2017\)
\(\Rightarrow B_{max}=2017\)
\("="\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy, GTLN của B là 2017 \(\Leftrightarrow x=\dfrac{1}{2}\)