a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)
\(\Leftrightarrow\left|3x-1\right|=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)
b)TH1: \(x\ge3\)
\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)
TH2: \(-5\le x< 3\)
\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)
TH3: \(x< -5\)
\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)
\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)
\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)
\(2.|3x-1|=\dfrac{4}{5}\)
\(|3x-1|=\dfrac{4}{5}:2\)
\(|3x-1|=\dfrac{2}{5}\)
\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)
\(3x-1=\dfrac{2}{5}\)
\(3x=\dfrac{2}{5}+1\)
\(3x=\dfrac{7}{5}\)
\(x=\dfrac{7}{5}:3\)
\(x=\dfrac{7}{15}\)
\(3x-1=-\dfrac{2}{5}\)
\(3x=-\dfrac{2}{5}+1\)
\(3x=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:3\)
\(x=\dfrac{1}{5}\)
\(a,\Rightarrow\left|3x-1\right|=\dfrac{4}{5}:2=\dfrac{2}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\left(x\ge\dfrac{1}{3}\right)\\1-3x=\dfrac{2}{5}\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{7}{5}\left(x\ge\dfrac{1}{3}\right)\\3x=\dfrac{3}{5}\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\left(tm\right)\\x=\dfrac{1}{5}\left(tm\right)\end{matrix}\right.\)
\(b,\Rightarrow\left[{}\begin{matrix}-x-5+3-x=9\left(x< -5\right)\\x+5+3-x=9\left(-5\le x< 3\right)\\x+5+x-3=9\left(x\ge3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\left(tm\right)\\0x=1\left(ktm\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)