Toán [Toán 9] Tìm GTLN

K

kienthuccohan

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V

vipboycodon

$P = \dfrac{a-1}{a}+\dfrac{b-1}{b}+\dfrac{c-4}{c}$

$= 3-(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{4}{c})$

$\le 3-\dfrac{(1+1+2)^2}{a+b+c} = \dfrac{1}{3}$

Dấu "=" xảy ra khi $a = b = \dfrac{3}{2}$, $c= 3$
 
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