Lời giải:

$-E=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

$\Rightarrow E\leq \frac{25}{12}$

Vậy $E_{\max}=\frac{25}{12}$. Giá trị này đạt được khi $x+\frac{1}{6}=0\Leftrightarrow x=\frac{-1}{6}$

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

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

\(\Rightarrow A=-\left(x^2+6x+9-9\right)+3\)

\(\Rightarrow A=-\left(x^2+6x+9\right)+3+9\)

\(\Rightarrow A=-\left(x+3\right)^2+12\le12\left(-\left(x+3\right)^2\le0,\forall x\right)\)

\(\Rightarrow Max\left(A\right)=12\left(tạix=-3\right)\)

Max(A)=12(tạix=−3)

\(D=-2x^2+3x-1\)

\(\Rightarrow D=-2\left(x^2-\dfrac{3}{2}x\right)-1\)

\(\Rightarrow D=-2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{9}{4}\right)-1\)

\(\Rightarrow D=-2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{4}\right)-1+\dfrac{9}{2}\)

\(\Rightarrow D=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{2}\le-\dfrac{7}{2}\left(-2\left(x-\dfrac{3}{2}\right)^2\le0,\forall x\right)\)

\(\Rightarrow Max\left(D\right)=-\dfrac{7}{2}\left(tạix=\dfrac{3}{2}\right)\)

MAX_D = -7/2 khi x = 3/2

\(C=-x^2-3x+4\)

\(\Rightarrow C=-\left(x^2+3x\right)+4\)

\(\Rightarrow C=-\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}\right)+4\)

\(\Rightarrow C=-\left(x^2+3x+\dfrac{9}{4}\right)+4+\dfrac{9}{4}\)

\(\Rightarrow C=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\left(-\left(x+\dfrac{3}{2}\right)^2\le0,\forall x\right)\)

\(\Rightarrow Max\left(C\right)=\dfrac{25}{4}\left(tạix=-\dfrac{3}{2}\right)\)

MAX_C = 25/4 khi x =-3/2

\(B=-x^2+8x-1=-\left(x^2-2.x.4+4^2\right)+15=-\left(x-4\right)^2+15\\ Vì:\left(x-4\right)^2\ge0\forall x\in R\\ \Rightarrow15-\left(x-4\right)^2\le15\forall x\in R\\ Vậy:max_B=15.khi.x=4\)

MAX_B = 15 khi x = 4

A = \(-x^2\)\(-6x+3\)

A = \(-x^2\)\(-6x+9-6\)

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

Dấu "=" xảy ra ⇔ \(x-3=0\)

                        ⇔ \(x=3\)

Vậy Amax =\(-6\) ⇔ \(x=3\)