a) \(\left|3x\right|=1-3x\)
TH1: \(3x>0\Rightarrow x>0\)
Ta có: \(3x=1-3x\)
\(\Rightarrow3x+3x=1\)
\(\Rightarrow6x=1\Rightarrow x=\frac{1}{6}\) (t/m)
TH2: \(3x< 0\Rightarrow x< 0\)
Lại có: \(3x=-1+3x\)
\(\Rightarrow3x-3x=-1\)
\(\Rightarrow0=-1\) (loại)
Vậy \(x=\frac{1}{6}.\)
b) \(\left|x-\frac{1}{5}\right|=\frac{1}{5}-x\)
\(\Rightarrow\left|x-\frac{1}{5}\right|=-\left(x-\frac{1}{5}\right)\)
\(\Rightarrow x-\frac{1}{5}\le0\)
\(\Rightarrow x\le0+\frac{1}{5}\)
\(\Rightarrow x\le\frac{1}{5}.\)
Vậy \(x\le\frac{1}{5}.\)
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