ta có:\(A=\frac{x^2-2x+2006}{x^2}=\frac{2006x^2-2.2006.x+2006^2}{2006x^2}\)
A=\(\frac{\left(x-2006\right)^2+2005x^2}{2006x^2}=\frac{\left(x-2006\right)^2}{2006x^2}+\frac{2005}{2006}\ge\frac{2005}{2006}\forall x\)
dấu = xảy ra khi x=2006
vậy Amin= 2005/2006 khi x=2006
dk:\(x\ne0\)
\(A=1-\frac{2}{x}+\frac{2006}{x^2}\)
đặt \(y=\frac{1}{x}\Rightarrow A=1-2y+2006y^2=2006\left(y^2-2.\frac{1}{2006}y+\frac{1}{2006^2}-\frac{1}{2006^2}\right)+1\)
\(A=2006\left(y-\frac{1}{2006}\right)^2-\frac{1.2006}{2006^2}+1=2006\left(y-\frac{1}{2006}\right)^2+\frac{2005}{2006}\)
\(\Rightarrow A\ge\frac{2005}{2006}\Rightarrow A_{min}=\frac{2005}{2006}\Leftrightarrow y=\frac{1}{2006}\)
từ đó thay y=\(\frac{1}{x}\) vào A là xong
A=\(\dfrac{x^2-2x+2016}{x^2}\)
<=>Ax2=x2-2x+2016
<=>(A-1)x2+2x-2016=0
\(\Delta\)=4-4(A-1)(-2016)\(\ge0\)
<=>4+4.2016A-4.2016\(\ge0\)
<=>A\(\ge\)\(\dfrac{4\left(2016-1\right)}{4.2016}=\dfrac{2015}{2016}\)
=>MinA =\(\dfrac{2015}{2016}\)khi x=2016
