Chứng minh đẳng thức sai

K

kakashi_hatake

$2 \le x \le 4 $
$\dfrac{1}{\sqrt{x-2}+1} <1<2 \ le x$
$\dfrac{1}{\sqrt{4-x}+1}>0$
-> $\dfrac{1}{\sqrt{x-2}+1} - \dfrac{1}{\sqrt{4-x}+1}-(x+\dfrac{1}{2}) <0$
đpcm
 
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