\(A-\frac{2013}{2014}=\frac{x^2-2x+2014}{x^2}-\frac{2013}{2014}=\frac{2014x^2-2.2014.x+2014^2-2013x^2}{2014x^2}\)
\(=\frac{x^2-2.x.2014+2014^2}{2014x^2}=\frac{\left(x-2014\right)^2}{2014x^2}\ge0\)
=>\(A\ge\frac{2013}{2014}\)
Dấu "=" xảy ra khi x=2014
Vậy minA=2013/2014 khi x=2014
A=\(\frac{2014x^2-2.2014x-2014^2}{2014x^2}\)=\(\frac{2013x^2+\left(x^2-2.2014x-2014^2\right)}{2014x^2}\)=\(\frac{2013x^2+\left(x-2014\right)^2}{2014x^2}\)=\(\frac{2013}{2014}+\frac{\left(x-2014\right)^2}{2014x^2}\ge\frac{2013}{2014}\)
vậy minA=\(\frac{2013}{2014}\)dấu bằng xảy ra khi x=2014
Tớ có cách khác hay nè :
\(A=\frac{x^2-2x+2014}{x^2}\)
\(A=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2014}{x^2}\)
\(A=1-\frac{2}{x}+\frac{2014}{x^2}\)
\(A=1-2\cdot\frac{1}{x}+2014\cdot\left(\frac{1}{x}\right)^2\)
Đặt \(a=\frac{1}{x}\)ta có :
\(A=1-2a+2014a^2\)
\(A=2014\cdot\left(a^2-\frac{1}{1007}a+\frac{1}{2014}\right)\)
\(A=2014\cdot\left[a^2-2\cdot a\cdot\frac{1}{2014}+\left(\frac{1}{2014}\right)^2+\frac{2013}{2014^2}\right]\)
\(A=2014\cdot\left[\left(a-\frac{1}{2014}\right)^2+\frac{2013}{2014^2}\right]\)
\(A=2014\cdot\left(a-\frac{1}{2014}\right)^2+\frac{2013}{2014}\ge\frac{2013}{2014}\forall a\)
Dấu "=" xảy ra \(\Leftrightarrow a=\frac{1}{2014}\)
\(\Leftrightarrow\frac{1}{x}=\frac{1}{2014}\)
\(\Leftrightarrow x=2014\)
Vậy Amin = \(\frac{2013}{2014}\)khi và chỉ khi x = 2014
