$1$) `1/{2x} + 4/5 = 14/{24x} - 3/2`
`⇔ 1/{2x} + 4/5 = 7/{12x} - 3/2`
`⇔ 7/{12x} - 1/{2x} = 4/5 + 3/2`
`⇔ {7-6}/{12x} = 4/5 + 3/2`
`⇔ 1/{12x} = {23}/{10}`
`⇒ 276x = 10`
`⇔ x = {10}/{276}=5/{138}`
Vậy `x=5/{138}`.
$2$) `-2|1/{2x} - 1/3| + 3/2 = 1/4`
`⇔ -2|1/{2x} - 1/3| = -5/4`
`⇔ |1/{2x} - 1/3| = 5/8`
`⇒` \(\left[ \begin{array}{l}\dfrac{1}{2x}-\dfrac{1}{3}=\dfrac{5}{8}\\\dfrac{1}{2x}-\dfrac{1}{3}=-\dfrac{5}{8}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}\dfrac{1}{2x} = \dfrac{23}{24}\\\dfrac{1}{2x}=\dfrac{-7}{24}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{12}{23}\\x=\dfrac{-12}{7}\end{array} \right.\)
Vậy $x$ $∈$ `{{-12}/7;{12}/{23}}`
$3$) `(x-3/7)^2 = 4`
`⇒` \(\left[ \begin{array}{l}x-\dfrac{3}{7}=2\\x-\dfrac{3}{7}=-2\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{17}{7}\\x=\dfrac{-11}{7}\end{array} \right.\)
Vậy $x$ $∈$ `{-{11}/7;{17}/7}`.
$4$) `{(x-2/3)}/{1/3} + 5/6 = 9 5/6`
`⇔{(x-2/3)}/{1/3} = 9`
`⇔ x - 2/3 = 3`
`⇔ x = {11}/3`
Vậy `x={11}/3`.
$5$) `(x-3/4)(x+2/5)=0`
`⇒` \(\left[ \begin{array}{l}x=\dfrac{3}{4}\\x=\dfrac{-2}{5}\end{array} \right.\)
Vậy $x$ $∈$ `{-2/5;3/4}`.
