Phải là tìm GTNN bạn nhé. Dấu của hạng tử bậc cao nhất là dương mà
\(A=\dfrac{x^2-2x+1995}{x^2}\\ =\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{1995}{x^2}\\ =1-\dfrac{2}{x}+\dfrac{1995}{x^2}\)
Đặt \(\dfrac{1}{x}=y\)
\(\Rightarrow A=1-2y+1995y^2\\ =1995y^2-2y+\dfrac{1}{1995}+\dfrac{1994}{1995}\\ =\left(1995y^2-2y+\dfrac{1}{1995}\right)+\dfrac{1994}{1995}\\ =1995\left(y^2-\dfrac{2}{1995}y+\dfrac{1}{1995^2}\right)+\dfrac{1994}{1995}\\ =1995\left(y-\dfrac{1}{1995}\right)^2+\dfrac{1994}{1995}\)
Do \(1995\left(y-\dfrac{1}{1995}\right)^2\ge0\forall x\)
\(\Rightarrow A=1995\left(y-\dfrac{1}{1995}\right)^2+\dfrac{1994}{1995}\ge\dfrac{1994}{1995}\forall x\)
Dấu "=" xảy ra khi:
\(1995\left(y-\dfrac{1}{1995}\right)^2=0\\ \Leftrightarrow y-\dfrac{1}{1995}=0\\ \Leftrightarrow y=\dfrac{1}{1995}\\ \Leftrightarrow\dfrac{1}{x}=\dfrac{1}{1995}\\ \Leftrightarrow x=1995\)
Vậy \(A_{\left(Min\right)}=\dfrac{1994}{1995}\) khi \(x=1995\)
