|2x-3|=x-1
TH1: 2x \(-\) 3 ≥ 0 <=> x ≥ 0 ta có:
2x \(-\) 3 = x \(-\) 1
<=> 2x \(-\) x = \(-\)1 \(+\) 3
<=> x = 2 (nhận)
TH2: 2x \(-\) 3 < 0 <=> x < 0 ta có:
\(-\)2x \(+\) 3 = x \(-\) 1
<=> \(-\)2x \(-\) x = \(-\)1 \(-\) 3
<=> \(-\)3x = \(-\)4
<=> \(-\)x = \(-\) \(\dfrac{4}{3}\)
<=> x = \(\dfrac{4}{3}\) (loại)
Vậy S={2}