`b)`

`1/3.( 2x - 5) = -2/3- 3/2`

`1/3. (2x - 5) = -13/6`

        `2x - 5    = -13/6: 1/3`

        `2x - 5    = -13/2`

        `2x          = -13/2+ 5`

        `2x          = 3/2`

          `x          = 3/2: 2`

          `x          = 3/4`

`c)`

   `lx- 1/2l= 3/2`

`⇒x- 1/2    = ±3/2`

`TH1:`

`x- 1/2= -3/2`

`x         = -3/2+ 1/2`

`x         =  -1`

`TH2:`

`x- 1/2 = 3/2`

`x         = 3/2+ 1/2`

`x          = 4/2= 2`

Vậy `x∈ {-1; 2}`

`d)`

`3/4- 2. l 2.x - 2/3 l = 2`

        `2. l2.x- 2/3l = 3/4- 2`

        `2. l2.x- 2/3l = -5/4`

           ` l2.x- 2/3l = -5/4: 2`

            `l2.x- 2/3l = -5/8`

`⇒ 2.x- 2/3= ±5/8`

 `TH1:`

`2.x - 2/3= -5/8`

`2.x         =-5/8+ 2/3`

`2.x         = 1/24`

   `x         = 1/24: 2`

   `x         = 1/48`

`TH2:`

`2.x - 2/3= 5/8`

`2.x         = 5/8+ 2/3`

`2.x         = 31/24`

   `x         = 31/24: 2`

   `x         = 31/48`

Vậy `x∈ {1/48; 31/48}`

Đáp án:

 c. \(\left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
b.\frac{1}{3}\left( {2x - 5} \right) =  - \frac{2}{3} - \frac{3}{2}\\
 \to \frac{1}{3}\left( {2x - 5} \right) = \frac{{ - 4 - 9}}{6} =  - \frac{{13}}{6}\\
 \to 2x - 5 =  - \frac{{13}}{6}:\frac{1}{3} = \frac{{ - 13}}{2}\\
 \to 2x = \frac{{ - 13}}{2} + 5 = \frac{{ - 13 + 10}}{2}\\
 \to 2x =  - \frac{3}{2}\\
 \to x =  - \frac{3}{4}\\
c.\left| {x - \frac{1}{2}} \right| = \frac{3}{2}\\
 \to \left[ \begin{array}{l}
x - \frac{1}{2} = \frac{3}{2}\\
x - \frac{1}{2} =  - \frac{3}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \frac{{3 + 1}}{2}\\
x = \frac{{ - 3 + 1}}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
d.2\left| {2x - \frac{2}{3}} \right| = \frac{3}{4} - 2 = \frac{{3 - 8}}{4}\\
 \to 2\left| {2x - \frac{2}{3}} \right| =  - \frac{5}{4}\\
 \to \left| {2x - \frac{2}{3}} \right| =  - \frac{5}{4}:2 =  - \frac{5}{8}\\
Do:\left| {2x - \frac{2}{3}} \right| \ge 0\forall x \in R
\end{array}\)

⇒ Phương trình vô nghiệm