\(a)A = x^2 - 20x + 101\)
\(= x^2 - 2.x.10 + 100 + 1\\
= (x - 10)^2 + 1 ≥1\)
Vậy \(min_A=1\Leftrightarrow x=10\)
\(b)B=x^2-x+1\\ =\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}+1\\ =\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(min_B=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
\(c)C=2x^2+2x+1=2\left(x^2+x+\dfrac{1}{2}\right)=2\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{4}\right]=2\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\right]=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)Vì: \(2\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
Dấu ''='' xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy\( min_C=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)
