Question

cho A=\(\dfrac{x}{\left(x+2004\right)^2}\)tìm x để A max

Answer 1

sửa đề x>0 .....

ta có \(\dfrac{1}{A}=\dfrac{\left(x+2004\right)^2}{x}\)

= \(\dfrac{x^2+2.2004.x+2004^2}{x}\)

= \(\dfrac{4.2004x+\left(x-2004\right)^2}{x}\)

= \(\dfrac{4.2004x}{x}+\dfrac{\left(x-2004\right)^2}{x}\)

= \(4.2004+\dfrac{\left(x-2004\right)^2}{x}\)

vì (x-2004)2 ≥ 0 ∀x

x> 0

=> \(\dfrac{\left(x-2004\right)^2}{x}\ge0\)

=> \(4.2004+\dfrac{\left(x-2004\right)^2}{x}\ge2004.4\)

=> \(\dfrac{1}{A}\ge8016\)

=> \(A\le\dfrac{1}{8016}\)

max A = \(\dfrac{1}{8016}\) dấu "=" xảy ra khi x-2004=0

x=2004

vậy maxA=\(\dfrac{1}{8016}\) khi x=2004