`A=-x^2+2x+10`

`=-(x^2-2x)+10`

`=-(x-1)^2+11<=11`

Dấu "=" xảy ra khi `x=1`.

`B=4x-2x^2+8`

`=-2(x^2-2x)+8`

`=-2(x^2-2x+1)+10`

`=-2(x-1)^2+10<=10`

Dấu "=" xảy ra khi `x=1`

`C=-x^2-x+1`

`=-(x^2+x)+1`

`=-(x^2+x+1/4)+1+1/4`

`=-(x+1/2)^2+5/4<=5/4`

Dấu "=" xảy ra khi `x=-1/2`

`D=-4x^2+6x+3`

`=-(4x^2-6x)+3`

`=-(4x^2-6x+9/4)+21/4`

`=-(2x-3/2)^2+21/4<=21/4`

Dấu "=' xảy ra khi `2x=3/2<=>x=3/4`

\(a,A=-x^2+2x+10=-x^2+2x-1+11=-\left(x^2-2x+1\right)+11\)

\(=11-\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=11-\left(x-1\right)^2\le11\)

Vậy MaxA = 11 <=> x = 1 .

\(b,B=-2x^2+4x-2+10=-2\left(x^2-2x+1\right)+10=10-2\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow B=10-2\left(x-1\right)^2\le10\)

Vậy MaxB = 10 <=> x = 1 .

\(c,C=-x^2-\dfrac{1}{2}.2.x-\dfrac{1}{4}+\dfrac{5}{4}=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\)

- Thấy : \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow C=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)

Vậy MaxC = 5/4 <=> x = -1/2 .

\(d,D=-4x^2+6x+3=-4x^2+2x.2.\dfrac{6}{4}-\dfrac{9}{4}+\dfrac{21}{4}=-\left(4x^2-6x+\dfrac{9}{4}\right)+\dfrac{21}{4}\)

\(=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\)

- Thấy : \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\le\dfrac{21}{4}\)

Vậy MaxD=21/4 <=> x = 3/4 .

Ta có \(A=3+\frac{1}{\left(x+1\right)^2+2}\).

A đạt giá trị lớn nhất khi \(\left(x+1\right)^2+2\) đạt giá trị nhỏ nhất.

Điều này xảy ra khi \(x=-1\) và khi đó \(A=\frac{7}{2}\).

Vậy giá trị lớn nhất của A là \(\frac{7}{2}\)

\(A=\frac{3x^2+6x+10}{x^2+2x+3}=\frac{3x^2+6x+9+1}{x^2+2x+3}=\frac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\frac{1}{x^2+2x+3}\)

       =\(\frac{1}{\left(x^2+2x+1\right)+2}\)\(=\frac{1}{\left(x+1\right)^2}+\frac{1}{2}\)

\(\Rightarrow\)MaxA=\(\frac{1}{2}\) khi x=-1

Chú ý:Max là giá trị lớn nhất nha bạn

\(A=\frac{3x^2+6x+10}{x^2+2x+3}=\frac{3x^2+6x+9}{x^2+2x+3}+\frac{1}{x^2+2x+3}\)

\(=3+\frac{1}{x^2+2x+3}=3+\frac{1}{\left(x+1\right)^2+2}\le3+\frac{1}{2}=\frac{7}{2}\)

Dấu "=" xảy ra <=> x=-1

Vậy GTLN của A=7/2 khi x=-1

Ta có: 

\(A=\frac{3x^2+6x+1}{x^2+2x+3}\)

   \(=\frac{3x^2+6x+9}{x^2+2x+3}+\frac{1}{x^2+2x+3}\)

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

    \(=3+\frac{1}{x^2+2x+3}\)

Lại có: \(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2\)

\(\Rightarrow\frac{1}{x^2+2x+3}\le\frac{1}{2}\)

\(\Rightarrow A\le3+\frac{1}{2}=\frac{7}{2}\)

Dấu = xảy ra khi \(x^2+2x+3=2\Rightarrow x=-1\)

Vậy \(A_{Min}=\frac{7}{2}\Leftrightarrow x=-1\)

\(A=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ A_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

\(A=\frac{2x^2+6x+10}{x^2+3x+3}=\frac{2\left(x^2+3x+3\right)+4}{x^2+3x+3}=2+\frac{4}{x^2+3x+3}\)

Để A đạt GTLN thì x²+3x+3 bé nhất

mà x²+3x+3=\(x^2+3.\frac{2}{3}x+\frac{2^2}{3^2}+\frac{23}{9}=\left(x+\frac{2}{3}\right)^2+\frac{23}{9}\ge\frac{23}{9}\)

Dấu "=" xảy ra khi \(x+\frac{2}{3}=0=>x=\frac{-2}{3}\)

lúc đó \(A=2+\frac{4}{\frac{23}{9}}=2+4.\frac{9}{23}=2+\frac{36}{23}=\frac{82}{23}\)

Vậy GTLN của \(A=\frac{82}{23}\)khi \(x=\frac{-2}{3}\)