\(a,-x-\dfrac{3}{7}=-\dfrac{7}{12}\)
`=>`\(-x=\dfrac{-7}{12}+\dfrac{3}{7}\)
`=>`\(-x=\dfrac{-49}{84}+\dfrac{36}{84}\)
`=>`\(-x=\dfrac{-13}{84}\)
`=>`\(x=\) `13/84`
Vậy ........
\(d,\dfrac{6}{5}-\left|x-1\right|=0,3\)
`=>`\(\left|x-1\right|=\dfrac{6}{5}-\dfrac{3}{10}\)
`=>`\(\left|x-1\right|=\dfrac{12}{10}-\dfrac{3}{10}\)
`=>`\(\left|x-1\right|=\dfrac{9}{10}\)
`=>`\(x-1=\dfrac{9}{10}\) hoặc \(x-1=\dfrac{-9}{10}\)
`=>`\(x=\dfrac{9}{10}+1\) \(x=\dfrac{-9}{10}+1\)
`=>`\(x=\dfrac{9}{10}+\dfrac{10}{10}\) \(x=\dfrac{-9}{10}+\dfrac{10}{10}\)
`=>`\(x=\dfrac{19}{10}\) \(x=\dfrac{1}{10}\)
a) \(-x-\dfrac{3}{7}=-\dfrac{7}{12}\Leftrightarrow-x=-\dfrac{13}{84}\Leftrightarrow x=\dfrac{13}{84}\)
b) \(\dfrac{6}{5}-\left|x-1\right|=0,3\Leftrightarrow-\left|x-1\right|=\dfrac{3}{10}-\dfrac{6}{5}\Leftrightarrow\left|x-1\right|=\dfrac{9}{10}\)
TH1: \(x-1=\dfrac{9}{10}\Leftrightarrow x=\dfrac{19}{10}\)
TH2: \(x-1=-\dfrac{9}{10}\Leftrightarrow x=\dfrac{1}{10}\)