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thaiha_98

a. $\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}=\sqrt{(x-3)^2}- \sqrt{(x+3)^2}=|x-3|-|x+3|$
 
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icy_tears

a, $\sqrt{x^2 - 6x + 9} - \sqrt{x^2 + 6x + 9}$
$= \sqrt{(x - 3)^2} - \sqrt{(x + 3)^2}$
$= |x - 3| - |x + 3|$

b, $A = 1$
\Leftrightarrow $|x - 3| - |x + 3| = 1$
Với $x < 3$ ta có: $3 - x - (x + 3) = -2x = 1 $ \Leftrightarrow $x = -0,5$
Với $x$ \geq $3$ ta có: $x - 3 - (x + 3) = -6$ khác 1
Vậy $x < 3$ thì $A = 1$
 
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