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

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

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

Dấu "=" xảy ra <=> x=-1

Vậy GTLN của A=7/2 khi x=-1

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

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 \(x=\frac{-2}{3}\)

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

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

Vì \(\left(x+1\right)^2\) lớn hơn hoặc bằng 0

=>\(\frac{1}{\left(x+1\right)^2+2}\) nhỏ hơn hoặc bằng \(\frac{1}{2}\)

=>A nhỏ hơn hoặc bằng 3+\(\frac{1}{2}\)=\(\frac{7}{2}\)

Dấu bằng xảy ra <=> x+1=0<=> x=-1

Vậy x=-1

\(A=\frac{3\left(2x^2+6x+10\right)}{3\left(x^2+3x+3\right)}=\frac{6x^2+18x+30}{3\left(x^2+3x+3\right)}=\frac{22\left(x^2+3x+3\right)-16x^2-48x-36}{3\left(x^2+3x+3\right)}\)

\(A=\frac{22}{3}-\frac{16x^2+48x+36}{3\left(x^2+3x+3\right)}=\frac{22}{3}-\frac{\left(4x+6\right)^2}{3\left(x^2+3x+3\right)}\)

Do \(\left\{{}\begin{matrix}\left(4x+6\right)^2\ge0\\x^2+3x+3=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\frac{\left(4x+6\right)^2}{3\left(x^2+3x+3\right)}\ge0\)

\(\Rightarrow A\le\frac{22}{3}\Rightarrow A_{max}=\frac{22}{3}\) khi \(4x+6=0\Rightarrow x=-\frac{3}{2}\)

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

Ta có : \(\left(x+1\right)^2+2\ge2\forall x\)

\(\Rightarrow\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)

\(\Rightarrow3+\frac{1}{\left(x+1\right)^2+2}\le\frac{7}{2}\forall x\) có GTLN là \(\frac{7}{2}\) tại x = - 1

Vậy .................

\(\frac{3x^2+6x+10}{x^2+2x+3}=\frac{3x^2+6x+6}{x^2+2x+3}+\frac{4}{x^2+2x+3}=\frac{3\left(x^2+2x+3\right)}{x^2+2x+3}+\frac{4}{\left(x^2+2x+1\right)+2}\)

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

Vì \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+2\ge2\Rightarrow\frac{4}{\left(x+1\right)^2+2}\le2\Rightarrow3+\frac{4}{\left(x+1\right)^2+2}\le5\)

=>giá trị nhỏ nhất của biểu thức là 5 <=>(x+1)²=0 <=. x+1=0 <=> x=-1

chị My nhầm rồi chị ơi

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

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

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

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

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

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

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

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