\(A=\frac{3x^2-18x+35}{x^2-6x+10}=\frac{3\left(x^2-6x+10\right)+5}{x^2-6x+10}=3+\frac{5}{\left(x-3\right)^2+1}\)
Ta có \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+1\ge1\Leftrightarrow\frac{5}{\left(x-3\right)^2+1}\le5\)
\(\Rightarrow A=3+\frac{5}{\left(x-3\right)^2+1}\le3+5=8\) có GTLN là 8
Dấu "=" xảy ra <=> x = 3
Vậy \(A_{max}=8\) tại \(x=3\)