1, -5

2, 0

3, 2011

4, ^o^

Michiel Girl Mít ướt

chuẩn

Biểu thức này bằng 0 bạn nhé!

GTNN = 0 (đúng rồi đấy )

Đặt

\(A=\dfrac{x^2-2x+2007}{2007x^2}=\dfrac{2007x^2-2\cdot x\cdot2007\cdot2007^2}{2007^2x^2}\)

\(\Rightarrow A=\dfrac{\left(x-2007\right)^2}{2007^2x^2}+\dfrac{2006}{2007^2}\ge\dfrac{2006}{2007^2}\)

Dấu ''='' xảy ra

\(\Leftrightarrow\dfrac{\left(x-2007\right)^2}{2007^2x^2}=0\Rightarrow\left(x-2007\right)^2=0\)

\(\Rightarrow x=2007\)

Vậy \(A_{MIN}=\dfrac{2006}{2007^2}\Leftrightarrow x=2007\)

Đặt A=\(\dfrac{x^2-2x+2007}{2007x^2}\)

2007A=\(\dfrac{2007x^2-2.2007x^2+2007^2}{2007x^2}\)

2007A-\(\dfrac{2006}{2007}\)=\(\dfrac{2007x^2-2.2007x+2007^2-2006x^2}{2007x^2}\)

2007A-\(\dfrac{2006}{2007}\)=\(\dfrac{x^2-2.2007x+2007^2}{2007x^2}\)

2007A-\(\dfrac{2006}{2007}\)=\(\dfrac{\left(x-2007\right)^2}{2007x^2}>=0\)

=>2007A>=\(\dfrac{2006}{2007}\)

=>A>=\(\dfrac{2006}{2007^2}\)

=>GTNN của A=\(\dfrac{2006}{2007^2}\)Dấu = xảy ra khi x=2007

\(B=\frac{\left(2007\cdot2009\right)+2007}{\left(2007\cdot2009\right)+2009}\)

\(B=\frac{2007}{2009}< 1\)

Vậy B<1

( chấm là nhân nhé)

nhớ ti ck đấy

\(A=\dfrac{x^2-2x+2007}{2007x^2}=\dfrac{2006}{2007^2}+\dfrac{x^2-4014x+2007^2}{2007^2x^2}=\dfrac{2006}{2007^2}+\dfrac{\left(x-2007\right)^2}{2007^2x^2}\ge\dfrac{2006}{2007^2}\)

Vậy GTNN là \(A=\dfrac{2006}{2007^2}\) đạt được khi \(x=2007\)