\(\left|x-3\right|+\left(x+3\right)=0\)

Case 1 :\(x-3\ge0\Leftrightarrow x\ge3\),so we have :

x - 3 + x + 3 = 0 => 2x = 0 => x = 0 (doesn't satisfy \(x\ge3\))

Case 2 :\(x-3< 0\Leftrightarrow x< 3\),so we have :

3 - x + x + 3 = 0 => 6 = 0 (absurd)

So there are no values of x

|x−3|+(x+3)=0

Case 1 :x−3≥0⇔x≥3

,so we have :

x - 3 + x + 3 = 0 => 2x = 0 => x = 0 (doesn't satisfy x≥3

)

Case 2 :x−3<0⇔x<3

,so we have :

3 - x + x + 3 = 0 => 6 = 0 (absurd)

So there are no values of x