\(a,A=x^2-5x+3\)
\(=\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{25}{4}+3\)
\(=\left[x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{13}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{13}{4}\) \(\ge-\dfrac{13}{4}\)
Dấu = xảy ra \(\Leftrightarrow x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(Min_A=-\dfrac{13}{4}\Leftrightarrow x=\dfrac{5}{2}\)
\(b,B=2x^2-7x+1\)
\(=2\left(x^2-\dfrac{7}{2}x+\dfrac{49}{16}\right)-\dfrac{49}{8}+1\)
\(=2\left(x-\dfrac{7}{4}\right)^2-\dfrac{41}{8}\ge-\dfrac{41}{8}\)
Dấu = xảy ra \(\text{⇔ }x-\dfrac{7}{4}=0\text{⇔ }x=\dfrac{7}{4}\)
Vậy \(Min_B=-\dfrac{41}{8}\text{⇔ }x=\dfrac{7}{4}\)
\(c,C=3x^2+2x\)
\(=3\left(x^2+\dfrac{2}{3}x+\dfrac{1}{3}\right)-\dfrac{1}{3}\)
\(=3\left[x^2+2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]-\dfrac{1}{3}\)
\(=3\left(x+\dfrac{1}{3}\right)^2-\dfrac{1}{3}\ge-\dfrac{1}{3}\)
Dấu = xảy ra \(\text{⇔ }x+\dfrac{1}{3}=0\text{⇔ }x=-\dfrac{1}{3}\)
Vậy \(Min_C=-\dfrac{1}{3}\text{⇔ }x=-\dfrac{1}{3}\)