\(B=\frac{x^2-2x+2018}{x^2}\)
\(\Rightarrow B=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2018}{x^2}\)
\(\Rightarrow B=1-\left(\frac{2}{x}-\frac{2018}{x^2}\right)\)
\(B=\frac{x^2-2x+2018}{x ^2}\)
\(\Rightarrow\)\(Bx^2=x^2-2x+2018\)
\(\Rightarrow\)\(\left(B-1\right)x^2+2x-2018=0\)
Để phương trình có nghiệm thì:
\(\Delta'=1-\left(B-1\right).\left(-2018\right)\)\(\ge0\)
\(\Leftrightarrow\)\(2018B-2017\ge0\)
\(\Leftrightarrow\) \(B\ge\frac{2017}{2018}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{-1}{B-1}=\frac{-1}{\frac{2017}{2018}-1}=2018\)
Vậy \(Min\)\(B=\frac{2017}{2018}\) \(\Leftrightarrow\)\(x=2018\)
p/s: tham khảo
