\(A=\frac{6x^2+12x+27}{3\left(x^2+2x+4\right)}=\frac{7\left(x^2+2x+4\right)-x^2-2x-1}{3\left(x^2+2x+4\right)}=\frac{7}{3}-\frac{\left(x+1\right)^2}{3\left(x+1\right)^2+9}\le\frac{7}{3}\)
\(\Rightarrow A_{max}=\frac{7}{3}\) khi \(x+1=0\Leftrightarrow x=-1\)
\(B=\frac{x^2-x+1}{x^2+x+1}=\frac{3\left(x^2+x+1\right)-2x^2-4x-2}{x^2+x+1}=3-\frac{2\left(x+1\right)^2}{x^2+x+1}=3-\frac{2\left(x+1\right)^2}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\le3\)
\(\Rightarrow B_{max}=3\) khi \(x+1=0\Rightarrow x=-1\)
\(C=\frac{2x^2-6x+3}{x^2-2x+1}=\frac{3\left(x^2-2x+1\right)+x^2}{x^2-2x+1}=3+\frac{x^2}{\left(x-1\right)^2}\ge3\)
\(C\) chỉ tồn tại min, ko tồn tại max
