Question

Tìm GTLN của A=1/(x^2-4x+9)

Answer 1

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

Do \(\left(x-2\right)^2\ge0\) ; \(\forall x\Rightarrow\left(x-2\right)^2+5\ge5\) ; \(\forall x\)

\(\Rightarrow A\le\dfrac{1}{5}\)

\(A_{max}=\dfrac{1}{5}\) khi \(x=2\)