Toán 9 [ nâng cao]

P

pinkylun

$A=\sqrt{(x-1)^2}+\sqrt{x+3)^2}+\sqrt{(x+2)^2}$

$A=|x-1|+|x+3|+|x+2|$

Ta có:

$|x-1|+|x+3| \ge |1-x+x+3|=4$

$|x+2| \ge 0$

$=>|x-1|+|x+3|+|x+2| \ge 4$

$<=>x=-2$
 
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