\(B=x^2-x+5=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+5\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)(dấu "='' xảy ra \(\Leftrightarrow x=\dfrac{1}{2})\)
Vậy \(B_{min}=\dfrac{19}{4}\) tại \(x=\dfrac{1}{2}\)
\(C=2x^2-x-1=2\left(x^2-\dfrac{1}{2}x-\dfrac{1}{2}\right)\)
\(=2\left(x^2-2.x.\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{2}\right)\)
\(=2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{16}\right]\)
\(=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\) (dấu "='' xảy ra \(\Leftrightarrow x=\dfrac{1}{2})\)
Vậy \(C_{min}=\dfrac{7}{8}\) tại \(x=\dfrac{1}{2}\)
\(D=3x^2+x+1=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{3}\right)\)
\(=3\left(x^2+2.x.\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{1}{36}+\dfrac{1}{3}\right)\)
\(=3\left[\left(x+\dfrac{1}{6}\right)^2+\dfrac{11}{36}\right]\)
\(=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{11}{12}\ge\dfrac{11}{12}\) (dấu "='' xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\))
Vậy \(D_{min}=\dfrac{11}{12}\) tại \(x=-\dfrac{1}{6}\)
