$a, A=2x+\dfrac{\sqrt[]{x^{2}-6x+9} }{x-3}$
$⇔A=2x+\dfrac{\sqrt[]{x-3^{2}}}{x-3}$
$⇔A=2x+\dfrac{|x-3|}{x-3}$
$TH_{1}:$
$x-3>0$
$⇒x>3$
$⇔|x-3|=x-3$
$⇔A=2x+\dfrac{x-3}{x-3}$
$⇔A=2x+1$
$TH_{2}:$
$x-3<0$
$⇒x<3$
$⇔|x-3|=-(x-3)$
$⇔A=2x+\dfrac{-(x-3)}{x-3}$
$⇔A=2x-1$
Vậy $A=2x-1$; $A=2x+1$
$b,$
Thay $x=-3$ (TMĐK) vào A ta có:
$A=2.(-3)+\dfrac{|-3-3|}{-3-3}$
$A=2.(-3)+\dfrac{6}{-6}$
$A=-7$
Vậy $A=-7$ khi $x=-3$
Xin hay nhất ạ
