\(\left|-x-\dfrac{4}{5}\right|=2\dfrac{1}{5}\)
\(\left|-x-\dfrac{4}{5}\right|=\dfrac{11}{5}\)
Có 2 TH:
- TH1: \(-x-\dfrac{4}{5}=\dfrac{11}{5}\)
\(\Rightarrow-x=3\Rightarrow x=-3\)
- TH2: \(-x-\dfrac{4}{5}=-\dfrac{11}{5}\)
\(\Rightarrow-x=-\dfrac{7}{5}\Rightarrow x=\dfrac{7}{5}\)
Vậy: \(x=\left\{-3;\dfrac{7}{5}\right\}\)
\(\left|-x-\dfrac{4}{5}\right|=2\dfrac{1}{5}\)
⇔ \(\left|-x-\dfrac{4}{5}\right|=\dfrac{11}{5}\)
⇒ \(\left[{}\begin{matrix}-x-\dfrac{4}{5}=\dfrac{11}{5}\\-x-\dfrac{4}{5}=-\dfrac{11}{5}\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-3\\x=\dfrac{7}{5}\end{matrix}\right.\)
