a, \(A=x^2+3x+2=x^2+\dfrac{3}{2}.x.2+\dfrac{9}{4}-\dfrac{1}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\)
Dấu " = " khi \(\left(x+\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{-3}{2}\)
Vậy \(MIN_A=\dfrac{-1}{4}\) khi \(x=\dfrac{-3}{2}\)
b, \(B=4x^2+4x+8=4\left(x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{7}{4}\right)\)
\(=4\left(x+\dfrac{1}{2}\right)^2+7\ge7\)
Dấu " = " khi \(4\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{-1}{2}\)
Vậy \(MIN_B=7\) khi \(x=\dfrac{-1}{2}\)
c, \(C=x^2-5x-3\)
\(=x^2-\dfrac{5}{2}.x.2+\dfrac{25}{4}-\dfrac{37}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{37}{4}\ge\dfrac{-37}{4}\)
Dấu " = " khi \(\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(MIN_C=\dfrac{-37}{4}\) khi \(x=\dfrac{5}{2}\)
