`1.`
`a, 2/3x + 4 = -12`
`=> 2/3x = -12 - 4`
`=> 2/3x = -16`
`=> x = -16 : 2/3`
`=> x = -24`
Vậy `x = -24`
`b, 3/4 + 1/4 : x = -3`
`=> 1/4 : x = -3 - 3/4`
`=> 1/4 : x = -15/4`
`=> x = 1/4 : (-15)/4`
`=> x = -1/15`
Vậy `x = -1/15`
`c, 2|2x - 3| = 1/2`
`=> |2x - 3| = 1/2 : 2`
`=> |2x - 3| = 1/4`
`⇒` \(\left[ \begin{array}{l}2x-3=\frac{1}{4}\\2x-3=\frac{-1}{4}\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}2x=\frac{1}{4}+3\\2x=\frac{-1}{4}+3\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}2x=\frac{13}{4}\\2x=\frac{11}{4}\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=\frac{13}{4}:2\\x=\frac{11}{4}:2\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=\frac{13}{8}\\x=\frac{11}{8}\end{array} \right.\)
Vậy`x in {13/8; 11/8}`
`d, |x + 4/15| - |-3,75| = -|-2,15|`
`=> |x + 4/15| - 3,75 = -2,15`
`=> |x + 4/15| = -2,15 + 3,75`
`=> |x + 4/15| = 8/5`
⇒ \(\left[ \begin{array}{l}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=\frac{8}{5}-\frac{4}{15}\\x=\frac{-8}{5}-\frac{4}{15}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=\frac{4}{3}\\x=\frac{-28}{15}\end{array} \right.\)
Vậy `x in {4/3; -28/15}`
