đề thi

V

vipboycodon

$A = |x-2|+3+x$
$= |2-x|+3+x$
$= 2-x+3+x$
$= 5$
Vậy $min A = 5$ khi $x \le 2$
 
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A

angleofdarkness

$ \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + .... + \dfrac{1}{\sqrt{100}}$

$>\dfrac{1}{\sqrt{100}} + \dfrac{1}{\sqrt{100}} + .... + \dfrac{1}{\sqrt{100}}$

$>\dfrac{1}{\sqrt{100}}.100$

$=10$

\Rightarrow đpcm
 
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