Giải:

a) \(D=-4x^2-3x+2\)

\(\Leftrightarrow D=-4x^2-3x-\dfrac{9}{16}+\dfrac{41}{16}\)

\(\Leftrightarrow D=\dfrac{41}{16}-\left(4x^2+3x+\dfrac{9}{16}\right)\)

\(\Leftrightarrow D=\dfrac{41}{16}-\left(2x+\dfrac{3}{4}\right)^2\le\dfrac{41}{16}\)

\(\Leftrightarrow D_{Max}=\dfrac{41}{16}\)

b) \(A=x^2+x+1\)

\(\Leftrightarrow A=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(\Leftrightarrow A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Leftrightarrow A_{Min}=\dfrac{3}{4}\)

c) \(B=4x^2-3x+2\)

\(\Leftrightarrow B=4x^2-3x+\dfrac{9}{16}+\dfrac{41}{16}\)

\(\Leftrightarrow B=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{41}{16}\ge\dfrac{41}{16}\)

\(\Leftrightarrow B_{Min}=\dfrac{41}{16}\)

Vậy ...