\(\left|x-1\right|+\left|x-3\right|\)
\(=\)\(\left|x-1\right|+\left|3-x\right|\) ≥ \(\left|x-1+3-x\right|=2\)
Dấu "=" ⇔ \(\left(x-1\right)\left(3-x\right)\) \(\text{≥}\) 0
⇔ \(1\text{≤}x\text{≤}3\)
\(\left|x-1\right|+\left|x-3\right|\)
\(TH1:\text{Nếu }x< -3\text{ thì }x-1< 0;x+3< 0\)
\(\Rightarrow\left|x-1\right|=-x+1;\left|x+3\right|=-x-3\)
\(\text{Pt(1) trở thành:}-x+1-x-3=4\)
\(\Leftrightarrow-2x=6\Leftrightarrow x=-3\text{(loại)}\)
\(TH2:\text{Nếu }-3\le x< 1\text{ thì }x-1< 0;x+3>0\)
\(\Rightarrow\left|x-1\right|=-x+1;\left|x+3\right|=x+3\)
\(\text{Pt(1) trở thành:}-x+1+x+3=4\)
\(\Leftrightarrow0x=0\text{(luôn đúng) }\)
\(\text{Kết hợp với điều kiện ta được:}\Rightarrow-3\le x< 1\)
\(TH3:\text{Nếu }x\ge1\text{ thì }x-1>0;x+3>0\)
\(\Rightarrow\left|x-1\right|=x-1;\left|x+3\right|=x+3\)
\(\text{Pt(1) trở thành:
}x-1+x+3=4\)
\(\Leftrightarrow2x=20\Leftrightarrow x=1\text{(t/m)}\)
\(\text{Vậy x nằm trong khoảng:}-3\le x\le1\)
