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 ...
D = -4x2-3x+2
=\(-4x-3x-\dfrac{9}{16}+\dfrac{41}{16}\)
=\(-4\left(x^2+\dfrac{3}{4}x+\dfrac{9}{64}\right)+\dfrac{41}{16}\)
= \(-4\left(x+\dfrac{3}{8}\right)^2+\dfrac{41}{16}\)
=> Max D= \(\dfrac{41}{16}khix=\dfrac{-3}{8}\)
