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
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