Question

Answer 1

\(B=\dfrac{x^2-x+5}{x^2-x+1}=1+\dfrac{4}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

\(\left(x-\dfrac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Rightarrow\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

\(\Rightarrow\dfrac{4}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{16}{3}\)

\(\Rightarrow B\le\dfrac{19}{3}\)

Dấu "=" xảy ra khi \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy giá trị lớn nhất của B là \(\dfrac{19}{3}\) khi \(x=\dfrac{1}{2}\).