Question

Tìm GTLN

\(\frac{1}{2.(x-1)^2+3}\)

Answer 1

Ta có:

(x - 1)² \(\ge\) 0

\(\Rightarrow\) 2(x - 1)² \(\ge\) 0

\(\Leftrightarrow\) 2(x - 1)² + 3 \(\ge\) 3

\(\Leftrightarrow\) \(\dfrac{1}{2\left(x-1\right)^2+3}\) \(\le\) \(\dfrac{1}{3}\)

Do đó \(\left(\dfrac{1}{2\left(x-1\right)^2+3}\right)_{max}\) = \(\dfrac{1}{3}\) khi (x - 1)² = 0

\(\Leftrightarrow\) x - 1 = 0

\(\Leftrightarrow\) x = 1

Vậy \(\left(\dfrac{1}{2\left(x-1\right)^2+3}\right)_{max}\) = \(\dfrac{1}{3}\) khi x = 1