Lời giải:
\(|-\frac{1}{2}-x|=\frac{1}{3}\Rightarrow \left[\begin{matrix}
\frac{-1}{2}-x=\frac{1}{3}\\
\frac{-1}{2}-x=\frac{-1}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix}
x=\frac{-5}{6}\\
x=\frac{-1}{6}\end{matrix}\right.\)
\(\left|-\frac{1}{2}-x\right|=\frac{1}{3}\)
=> \(\left[{}\begin{matrix}-\frac{1}{2}-x=\frac{1}{3}\\-\frac{1}{2}-x=-\frac{1}{3}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\left(-\frac{1}{2}\right)-\frac{1}{3}\\x=\left(-\frac{1}{2}\right)-\left(-\frac{1}{3}\right)\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{5}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{5}{6};-\frac{1}{6}\right\}.\)
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∣∣−12−x∣∣=13|−12−x|=13
=> [−12−x=13−12−x=−13[−12−x=13−12−x=−13 => ⎡⎢⎣x=(−12)−13x=(−12)−(−13)[x=(−12)−13x=(−12)−(−13)
=> [x=−56x=−16[x=−56x=−16
Vậy x∈{−56;−16}.x∈{−56;−16}.
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