\(B=\dfrac{1}{4x^2-4x+5}\)
Ta có : \(4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\)
Nhận xét :
\(\left(2x-1\right)^2\ge0\)
\(\Leftrightarrow\left(2x-1\right)^2+4\ge4\)
\(\Leftrightarrow\dfrac{1}{\left(2x-1\right)^2+4}\le\dfrac{1}{4}\)
\(\Leftrightarrow B\le\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)
Vậy \(B_{Max}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)