TH1: \(5x+3\ge0=>x\ge-\dfrac{3}{5}\)
Pt trở thành \(5x+3=x+5\)
<=> 4x = 2
<=> x = \(\dfrac{1}{2}\) (thoả)
TH2: \(5x+3< 0=>5x< -3=>x< -\dfrac{3}{5}\)
Pt trở thành 5x + 3 = -x - 5
6x = -8
x = \(-\dfrac{4}{3}\)(thoả)
Vậy \(x\in\left\{\dfrac{1}{2};-\dfrac{4}{3}\right\}\)
X = \(-\) 1 \(\dfrac{1}{3}\)
X = \(\dfrac{1}{2}\)