Question

    Tìm x để B có giá trị nhỏ nhất :  \(B=\frac{x^2-2x+2011}{x^2}\)với x > 0

Answer 1

Ta có :

 \(B=\frac{x^2-2x+2011}{x^2}\)

\(B=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2011}{x^2}\)

\(B=1-\frac{2}{x}+\frac{2011}{x^2}\)

\(B=\left(\frac{\sqrt{2011}^2}{x^2}-\frac{2}{x}+\frac{1}{2011}\right)+\frac{2010}{2011}\)

\(B=\left(\frac{\sqrt{2011}}{x}-\frac{1}{\sqrt{2011}}\right)^2+\frac{2010}{2011}\)

Mà : \(\left(\frac{\sqrt{2011}}{x}-\frac{1}{\sqrt{2011}}\right)^2\ge0\forall x\)

\(\Rightarrow B\ge\frac{2010}{2011}\)

Dấu "=" xảy ra khi :

\(\frac{\sqrt{2011}}{x}-\frac{1}{\sqrt{2011}}=0\)

\(\Leftrightarrow x=2\sqrt{2011}\)

Vậy \(MinB=\frac{2010}{2011}\Leftrightarrow x=2\sqrt{2011}\)