Question

Tìm GTNN: B=\(\dfrac{1}{2\left(x-1\right)^2+3}\)

Answer 1

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

Xét \(2\left(x-1\right)^2\ge0\)

\(\Rightarrow\)\(2\left(x-1\right)^2+3\ge3\)

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

Vậy GTLN của B là \(\dfrac{1}{3}\) \(\Leftrightarrow x=1\)

Answer 2

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

\(2\left(x-1\right)^2\ge0\)

Suy ra:

\(2\left(x-1\right)^2+3\ge3\)

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

Dấu "=" xảy ra khi:

\(x=1\)