Question

/ x - \(\dfrac{1}{5}\)/ - x = \(\dfrac{1}{3}\)

Answer 1

\(\left|x-\dfrac{1}{5}\right|-x=\dfrac{1}{3}\)

\(\Rightarrow\left|x-\dfrac{1}{5}\right|=x+\dfrac{1}{3}\)

\(\Rightarrow x-\dfrac{1}{5}=x+\dfrac{1}{3}\) hoặc \(\Rightarrow x-\dfrac{1}{5}=-x-\dfrac{1}{3}\)

\(\Rightarrow-\dfrac{1}{5}=\dfrac{1}{3}\) (loại) hoặc \(2x=-\dfrac{2}{15}\Rightarrow x=-\dfrac{1}{15}\)

Answer 2

+) Xét \(x\ge\dfrac{1}{5}\) ta có:

\(x-\dfrac{1}{5}-x=\dfrac{1}{3}\Rightarrow\dfrac{-1}{5}=\dfrac{1}{3}\) ( vô lí )

+) Xét \(x< \dfrac{1}{5}\) ta có:

\(\dfrac{1}{5}-x-x=\dfrac{1}{3}\)

\(\Rightarrow2x=\dfrac{-2}{15}\)

\(\Rightarrow x=\dfrac{-1}{15}\) ( t/m )

Vậy \(x=\dfrac{-1}{15}\)

Answer 3

Ta có: \(\left|x-\dfrac{1}{5}\right|-x=\dfrac{1}{3}\)

\(\Rightarrow\left|x-\dfrac{1}{5}\right|=\dfrac{1}{3}+x\)

+) \(TH1:x-\dfrac{1}{5}\ge0\Rightarrow x\ge\dfrac{1}{5}\)

Khi đó: \(x-\dfrac{1}{5}=\dfrac{1}{3}+x\)

\(\Rightarrow x-x=\dfrac{1}{5}+\dfrac{1}{3}\)

\(\Rightarrow0=\dfrac{8}{15}\) (loại)

+) \(TH2:x-\dfrac{1}{5}< 0\Rightarrow x< \dfrac{1}{5}\)

Khi đó: \(-x+\dfrac{1}{5}=\dfrac{1}{3}+x\)

\(\Rightarrow-x-x=\dfrac{-1}{5}+\dfrac{1}{3}\)

\(\Rightarrow-2x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-1}{15}\left(tm\right)\)

Vậy \(x=\dfrac{-1}{15}.\)