Ta có:
\(B=\dfrac{x^2-2x+2016}{x^2}\left(x\ne0\right)\)
\(B=\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{2016}{x^2}\)
\(B=1-\dfrac{2}{x}+\dfrac{2016}{x^2}\)
\(B=2016\left(\dfrac{1}{x^2}-\dfrac{1}{2016}\cdot\dfrac{2}{x}+\dfrac{1}{2016}\right)\)
\(B=2016\cdot\left(\dfrac{1}{x^2}-2\cdot\dfrac{1}{2016}\cdot\dfrac{1}{x}+\dfrac{1}{2016}\right)\)
\(B=2016\left(\dfrac{1}{x^2}-2\cdot\dfrac{1}{2016}\cdot\dfrac{1}{x}+\dfrac{1}{4064256}+\dfrac{2015}{4064256}\right)\)
\(B=2016\left[\dfrac{1}{x^2}-2\cdot\dfrac{1}{2016}\cdot\dfrac{1}{x}+\dfrac{1}{4064256}\right]+2016\cdot\dfrac{2015}{4064256}\)
\(B=2016\cdot\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2+\dfrac{2015}{2016}\)
Ta có: \(2016\cdot\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2\ge0\forall x\)
\(\Rightarrow2016\cdot\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2+\dfrac{2015}{2016}\ge\dfrac{2015}{2016}\forall x\)
Dấu "=" xảy ra khi:
\(2016\cdot\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2+\dfrac{2015}{2016}=\dfrac{2015}{2016}\)
\(\Leftrightarrow2016\cdot\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2=0\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{2016}=0\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{2016}\)
\(\Leftrightarrow x=2016\left(tm\right)\)
Vậy: \(B_{min}=\dfrac{2015}{2016}\Leftrightarrow x=2016\)
