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 ...

sao ra 9/16

\(E=3x^2-5x+1=3\left(x-\frac{5}{6}\right)^2-\frac{13}{12}\ge-\frac{13}{12}\)

Vậy Min E = -13/12 <=> x = 5/6

\(\frac{4x+1}{x^2+3}=\frac{x^2+4x+4-\left(x^2+3\right)}{x^2+3}=\frac{\left(x+2\right)^2}{x^2+3}-1\ge-1\)

Dấu "=' xảy ra khi x = -2

\(A=x^2-6x+3\)

\(=\left(x^2-6x+9\right)-6\)

\(=\left(x+3\right)^2-6\)

ma \(\left(x+3\right)^2\ge0\Leftrightarrow\left(x+3\right)^2-6\ge-6\)

vậy gtnn của A là -6 tại x=-3

\(B=x^2+3x+7=\left(x^2+2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{17}{4}\)

\(=\left(x+\frac{3}{2}\right)^2+\frac{17}{4}\ge\frac{17}{4}\)

vay .............................................

2/

\(A=-x^2+4x+8=-\left(x^2-4x+4\right)+12=-\left(x-2\right)^2+12\le12\)

vay .........................................

\(B=-x^2+3x-5=-\left(x^2-2\frac{3}{2}x+\frac{9}{4}\right)-\frac{11}{4}=\left(x-\frac{3}{2}\right)^2-\frac{11}{4}\le-\frac{11}{4}\)

vay.....................................

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ko sao cảm ơn