\(\left|2x-3\right|=x-1\) (ĐK: \(x\ge1\))
TH1: Nếu: \(2x-3\ge0\Leftrightarrow x\ge\dfrac{3}{2}\) thì:
\(2x-3=x-1\left(đk:x\ge\dfrac{3}{2}\right)\)
\(\Leftrightarrow2x-x=-1+3\)
\(\Leftrightarrow x=2\left(tm\right)\)
TH2: Nếu: \(2x-3< 0\Leftrightarrow x< \dfrac{3}{2}\) thì:
\(-\left(2x-3\right)=x-1\left(đk:1\le x< \dfrac{3}{2}\right)\)
\(\Leftrightarrow-2x+3=x-1\)
\(\Leftrightarrow-2x-x=-1-3\)
\(\Leftrightarrow-3x=-4\)
\(\Leftrightarrow x=\dfrac{4}{3}\left(tm\right)\)
Vậy: \(x\in\left\{2;\dfrac{4}{3}\right\}\)
|2x-3|=x-1 ( ĐKXĐ: x≥ 1). TH1 : 2x-3= x-1. <=> 2x-x=-1+3 <=> x = 2 ( T/m). TH2 : 2x-3= - (x-1) <=> 2x-3= -x+1. <=> 2x+x=1+3. <=> 3x=4. <=> x= 4/3 (T/m).
