\(C=x^2-3x+5\)
\(=x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\)
Vì \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
\(\Rightarrow C\ge\dfrac{11}{4}\forall x\)
Dấu "=" xảy ra khi \(\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(MIN_C=\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{2}.\)
\(D=3x^2-6x-1\)
\(=3\left(x^2-3x-\dfrac{1}{3}\right)\)
\(=3\left(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{31}{12}\right)\)
\(=3\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{31}{12}\right]\)
\(=3\left(x-\dfrac{3}{2}\right)^2-\dfrac{31}{4}\)
.......
Vậy \(MIN_D=\dfrac{-31}{4}\) khi \(x=\dfrac{3}{2}.\)
\(E=2x^2-6x\)
\(=2\left(x^2-3x\right)\)
\(=2\left[\left(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{9}{4}\right]\)
\(=2\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\)
\(=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
.....
Vậy \(MIN_E=\dfrac{-9}{2}\) khi \(x=\dfrac{3}{2}.\)
