Question

Cho 2 < x < 3. Tìm GTNN của \(P=\dfrac{1}{\left(x-2\right)^2}+\dfrac{1}{\left(3-x\right)^2}+\dfrac{1}{\left(x-2\right)\left(3-x\right)}\)

Answer 1

\(P\ge\dfrac{1}{2}\left(\dfrac{1}{x-2}+\dfrac{1}{3-x}\right)^2+\dfrac{4}{\left(x-2+3-x\right)^2}=\dfrac{1}{2}\left(\dfrac{1}{x-2}+\dfrac{1}{3-x}\right)^2+4\)

\(P\ge\dfrac{1}{2}\left(\dfrac{4}{x-4+3-x}\right)^2+4=12\)

Dấu "=" xảy ra khi \(x-2=3-x\Rightarrow x=\dfrac{5}{2}\)