Ta có:
\(\left|x\right|=x\) khi \(x\ge0\)
\(\left|x\right|=-x\) khi \(x< 0\)
\(\left|x-2\right|=x-2\) khi \(x\ge2\)
\(\left|x-2\right|=2-x\) khi \(x< 2\)
TH1: \(x< 0\)
\(\Rightarrow\left|x-\left(-x\right)\right|+3\left(2-x\right)+x=11\)
\(\Leftrightarrow\left|2x\right|+6-3x+x=11\)
\(\Leftrightarrow-2x+6-3x+x=11\)
\(\Leftrightarrow6-4x=11\)
\(\Leftrightarrow-4x=5\)
\(\Leftrightarrow x=\frac{-5}{4}< 0\) (chọn)
TH2: \(0\le x< 2\)
\(\Rightarrow\left|x-x\right|+3\left(2-x\right)+x=11\)
\(\Leftrightarrow6-3x+x=11\)
\(\Leftrightarrow6-2x=11\)
\(\Leftrightarrow-2x=5\)
\(\Leftrightarrow x=\frac{-5}{2}< 0\) (loại)
TH3: \(x\ge2\)
\(\Rightarrow\left|x-x\right|+3\left(x-2\right)+x=11\)
\(\Leftrightarrow3x-6+x=11\)
\(\Leftrightarrow4x-6=11\)
\(\Leftrightarrow4x=17\)
\(\Leftrightarrow x=\frac{17}{4}>2\) (chọn)
Vậy phương trình có tập nghiệm S={\(\frac{-5}{4};\frac{17}{4}\)}
