help me, please!!!!!!!!!!!!!!!
\(B=\frac{x^2-2x+2018}{2018x^2}\)
\(=\frac{1}{2018}-\frac{2}{2018x}+\frac{1}{x^2}\)
\(=\left(\frac{1}{x}-\frac{1}{\sqrt{2018}}\right)^2\ge0\)
Vậy giá trị nhỏ nhất \(B=0\)khi và chỉ khi \(\frac{1}{x}-\frac{1}{\sqrt{2018}}=0\)
\(\Rightarrow\frac{1}{x}=\frac{1}{\sqrt{2018}}\)
\(\Rightarrow x=\sqrt{2018}\)
\(B=\frac{x^2-2x+2018}{2018x^2}\)
\(=\frac{2018x^2-2\cdot2018\cdot x+2018^2}{2018^2x^2}\)
\(=\frac{2017x^2+\left(x^2-2\cdot2018\cdot x+2018^2\right)}{2018^2x^2}\)
\(=\frac{2017}{2018^2}+\frac{\left(x-2018\right)^2}{2018^2x^2}\ge\frac{2017}{2018^2}\)
Đẳng thức xảy ra tại x=2018
Vậy .........
