[TOán 9] Tìm MIN

V

vipboycodon

$\dfrac{\sqrt{x}+1}{\sqrt{x}-3} = \dfrac{\sqrt{x}-3+4}{\sqrt{x}-3} = 1+\dfrac{4}{\sqrt{x}-3} \ge \dfrac{-1}{3}$
Vậy $min = \dfrac{-1}{3}$ khi $x = 0.$
 
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