Question

tìm x>0 để \(B=\frac{x^2-2x+2011}{x^2}\) đạt giá trị nhỏ nhất

Answer 1

Ta có \(A=\frac{x^2-2x+2011}{x^2}\)

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

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

Đặt \(\frac{1}{x}=y\)ta có:

\(A=1-2y+2011y^2\)

\(A=2011y^2-2y+1\)

\(A=2011\left(y^2-\frac{2}{2011}y+\frac{2}{2011}\right)\)

\(=2011\left(y^2-2\times y\times\frac{1}{2011}+\frac{1}{2011^2}-\frac{1}{2011^2}+\frac{1}{2011}\right)\)

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

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

Vì (y-\(\frac{1}{2011}\))\(^2\)>=0

\(\Rightarrow2011\left(y-\frac{1}{2011}\right)^2+\frac{2010}{2011}\)

Hay \(A>=\frac{2010}{2011}\)

Answer 2

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