Ta có: \(\frac{1}{x^2-12x+2019}=\frac{1}{x^2-12x+36+1983}=\frac{1}{\left(x-6\right)^2+1983}\le\frac{1}{1983}\forall x\)
Dấu "=" xảy ra <=> x - 6 = 0
<=> x = 6
Vậy Max của \(\frac{1}{x^2-12x+2019}\)= 1983 <=> x = 6
\(x^2-12x+2019=\left(x^2-2\times x\times6+6^2\right)+1983=\left(x-6\right)^2+1983\ge1983\)
(vì \(\left(x-6\right)^2\ge0\Rightarrow\left(x-6\right)^2+1983\ge1983\))
\(\Rightarrow\frac{1}{\left(x-6\right)^2+1983}\le\frac{1}{1983}\)hay \(\frac{1}{x^2-12x+2019}\le\frac{1}{1983}\)
Dấu = xảy ra \(\Leftrightarrow\left(x-6\right)^2=0\Leftrightarrow x-6=0\Leftrightarrow x=6\)
Vậy GTLN của \(\frac{1}{x^2-12x+2019}\)là 1/1983
