Question

tìm GTNN của biểu thức A = \(\frac{x^2-2x+2006}{x^2}\)

Answer 1

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

 

Answer 2

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

Answer 3

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