a, \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=1\dfrac{2}{5}\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{10}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{10}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{10}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{11}{15}\\x=\dfrac{31}{15}\end{matrix}\right.\)
b, \(\left|\dfrac{1}{4}x-2\dfrac{1}{5}\right|=\left|0,6-\dfrac{2}{3}x\right|\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{4}x-2\dfrac{1}{5}=\dfrac{2}{3}x-0,6\\\dfrac{1}{4}x-2\dfrac{1}{5}=0,6-\dfrac{2}{3}x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}x=-0,6+2\dfrac{1}{5}\\\dfrac{1}{4}x+\dfrac{2}{3}x=0,6+2\dfrac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-\dfrac{5}{12}x=1,6\\\dfrac{11}{12}x=2,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3,84\\x=\dfrac{168}{55}\end{matrix}\right.\)
c, \(\left|2x-\dfrac{1}{3}\right|=x+\dfrac{1}{2}\)
+, Xét \(x\ge\dfrac{1}{6}\) thì \(2x-\dfrac{1}{3}\ge0\Rightarrow\left|2x-\dfrac{1}{3}\right|=2x-\dfrac{1}{3}\)
Thay vào ta có:
\(2x-\dfrac{1}{3}=x+\dfrac{1}{2}\Rightarrow x=\dfrac{5}{6}\)(chọn vì thoả mãn điều kiện \(x\ge\dfrac{1}{6}\))
+, Xét \(x< \dfrac{1}{6}\) thì \(2x-\dfrac{1}{3}< 0\Rightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{3}-2x\)
Thay vào ta có:
\(\dfrac{1}{3}-2x=x+\dfrac{1}{2}\Rightarrow3x=-\dfrac{1}{6}\Rightarrow x=-\dfrac{1}{18}\)(chọn vì thoả mãn điều kiện \(x< \dfrac{1}{6}\))
Vậy.............
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