\(\left|x+\dfrac{1}{3}\right|-\dfrac{1}{2}x=\dfrac{7}{6}\\ \left|x+\dfrac{1}{3}\right|=\dfrac{7}{6}-\dfrac{1}{2}x\)
Vì \(\left|x+\dfrac{1}{3}\right|\ge0\forall x\Rightarrow\dfrac{7}{6}-\dfrac{1}{2}x\ge0\forall x\\ \Leftrightarrow\dfrac{1}{2}x\le\dfrac{7}{6}\\ \Leftrightarrow x\le\dfrac{7}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{6}-\dfrac{1}{2}x\\x+\dfrac{1}{3}=-\dfrac{7}{6}+\dfrac{1}{2}x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}x=\dfrac{7}{6}-\dfrac{1}{3}\\x-\dfrac{1}{2}x=-\dfrac{7}{6}-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x=\dfrac{5}{6}\\\dfrac{1}{2}x=-\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{9}\\x=-3\end{matrix}\right.\)
Vì x phải thỏa mãn đk x bé hơn hoặc bằng \(\dfrac{7}{6}\) nên x chỉ có thể bằng \(\dfrac{5}{9}\).
Vậy x = \(\dfrac{5}{9}\)
\(|x+\dfrac{1}{3}|-\dfrac{1}{2}x=\dfrac{7}{6}\)
\(\Rightarrow|x+\dfrac{1}{3}|=\dfrac{7}{6}+\dfrac{1}{2}x\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{6}+\dfrac{1}{2}x\\x+\dfrac{1}{3}=-\dfrac{7}{6}-\dfrac{1}{2}x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}x=\dfrac{7}{6}-\dfrac{1}{3}\\x+\dfrac{1}{2}x=-\dfrac{7}{6}-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{5}{6}\\\dfrac{3}{2}x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}:\dfrac{1}{2}=\dfrac{5}{6}.\dfrac{1}{2}=\dfrac{5}{12}\\x=-\dfrac{3}{2}:\dfrac{3}{2}=-\dfrac{3}{2}.\dfrac{2}{3}=-1\end{matrix}\right.\)
Vậy x = 5/12 ; x = -1
