(1-2x)2 - 3|2x-1| = 0
=> (1-2x)(1-2x) - 3|2x-1|=0
=> (1-2x)-(2x-4x) - 3|2x-1| = 0
=> 1 - 3|2x-1| = 0
=> 3|2x-1|=1
=> |2x-1|= 1/3
=> 2x-1 \(\in\){1/3; -1/3}
=> x \(\in\){2/3; 1/3}
\(\left(1-2x\right)^2=3\left|2x-1\right|\)
TH1: \(2x-1< 0\)⇔\(x< \frac{1}{2}\)
⇒ \(\left|2x-1\right|=1-2x\)
⇒ \(\left(1-2x\right)^2=3\left(1-2x\right)\)
\(1-2x=3\)
\(x=\frac{1-3}{2}=-1\)( thỏa mãn)
TH2:\(\) 2x-1 ≥ 0 ⇔ x ≥ \(\frac{1}{2}\)
⇒\(\left|2x-1\right|=2x-1\)
\(\left(1-2x\right)^2=\left(2x-1\right)^2\)
\(\left(2x-1\right)^2=3\left(2x-1\right)\)
\(2x-1=3\)
\(x=\frac{3+1}{2}=\frac{4}{2}=2\)(thỏa mãn)
Vậy x=2 hoặc -1
