Giải:
`a, |x + 1/3| - 5 = 7`
`-> |x + 1/3| = 7 + 5`
`-> |x + 1/3| = 12`
`->` \(\left[ \begin{array}{l}x+\dfrac{1}{3} = 12\\x+\dfrac{1}{3} = -12\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x = 12 - \dfrac{1}{3} = \dfrac{36}{3} - \dfrac{1}{3} = \dfrac{35}{3}\\x = - 12 - \dfrac{1}{3} = \dfrac{-36}{3} - \dfrac{1}{3} = \dfrac{-37}{3}\end{array} \right.\)
Vậy `x ∈ { 35/3 ; -37/3 }`.
`b, 3(x - 2) + 2/5 = 4`
`-> 3(x - 2) = 4 - 2/5`
`-> 3(x - 2) = 20/5 - 2/5`
`-> 3(x - 2) = 18/5`
`-> x - 2 = 18/5 : 3`
`-> x - 2 = 18/5 . 1/3`
`-> x - 2 = 6/5`
`-> x = 6/5 + 2`
`-> x = 6/5 + 10/5`
`-> x = 16/5`
Vậy `x = 16/5`.
