$\\$

`a,`

`|2x-5|=4`

`->` \(\left[ \begin{array}{l}2x-5=4\\2x-5=-4\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}2x=9\\2x=1\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{9}{2}\\x=\dfrac{1}{2}\end{array} \right.\)

Vậy `x=9/2` hoặc `x=1/2`

$\\$

`b,`

`1/3 - |5/4 - 2x|=  1/4`

`-> |5/4 - 2x|=1/3 - 1/4`

`-> |5/4 - 2x|=1/12`

`->` \(\left[ \begin{array}{l}\dfrac{5}{4}-2x=\dfrac{1}{12}\\\dfrac{5}{4}-2x=\dfrac{-1}{12}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}2x=\dfrac{7}{6}\\2x=\dfrac{4}{3}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{7}{12}\\x=\dfrac{2}{3}\end{array} \right.\) 

Vậy `x=7/12` hoặc `x=2/3`

$\\$

`c,`

`|x-3/4| - 1/4=0`

`-> |x-3/4|=0+1/4`

`-> |x-3/4|=1/4`

`->` \(\left[ \begin{array}{l}x-\dfrac{3}{4}=\dfrac{1}{4}\\x-\dfrac{3}{4}=\dfrac{-1}{4}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=1\\x=\dfrac{1}{2}\end{array} \right.\) 

Vậy `x=1` hoặc `x=1/2`

$\\$

`d,`

`7,5 - 3 |5-2x|=-4,5`

`-> 3 |5-2x|=7,5 + 4,5`

`-> 3 |5-2x|=12`

`-> |5-2x|=12 : 3`

`-> |5-2x|=4`

`->` \(\left[ \begin{array}{l}5-2x=4\\5-2x=-4\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}2x=1\\2x=9\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{array} \right.\) 

Vậy `x=1/2` hoặc `x=9/2`

Đáp án:

\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = \dfrac{9}{2}\\
x = \dfrac{1}{2}
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = \dfrac{7}{{12}}\\
x = \dfrac{2}{3}
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{2}
\end{array} \right.\\
d)\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{9}{2}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\left| {2x - 5} \right| = 4\\
 \to \left[ \begin{array}{l}
2x - 5 = 4\left( {DK:x \ge \dfrac{5}{2}} \right)\\
2x - 5 =  - 4\left( {DK:x < \dfrac{5}{2}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{9}{2}\\
x = \dfrac{1}{2}
\end{array} \right.\\
b)\dfrac{1}{3} - \left| {\dfrac{5}{4} - 2x} \right| = \dfrac{1}{4}\\
 \to \left| {\dfrac{5}{4} - 2x} \right| = \dfrac{1}{{12}}\\
 \to \left[ \begin{array}{l}
\dfrac{5}{4} - 2x = \dfrac{1}{{12}}\left( {DK:\dfrac{5}{8} \ge x} \right)\\
\dfrac{5}{4} - 2x =  - \dfrac{1}{{12}}\left( {DK:\dfrac{5}{8} < x} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{7}{6}\\
2x = \dfrac{4}{3}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{7}{{12}}\\
x = \dfrac{2}{3}
\end{array} \right.\\
c)\left| {x - \dfrac{3}{4}} \right| = \dfrac{1}{4}\\
 \to \left[ \begin{array}{l}
x - \dfrac{3}{4} = \dfrac{1}{4}\left( {DK:x \ge \dfrac{3}{4}} \right)\\
x - \dfrac{3}{4} =  - \dfrac{1}{4}\left( {DK:x < \dfrac{3}{4}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{2}
\end{array} \right.\\
d)\dfrac{{15}}{2} - 3.\left| {5 - 2x} \right| =  - \dfrac{9}{2}\\
 \to 3.\left| {5 - 2x} \right| = 12\\
 \to \left| {5 - 2x} \right| = 4\\
 \to \left[ \begin{array}{l}
5 - 2x = 4\left( {DK:\dfrac{5}{2} \ge x} \right)\\
5 - 2x =  - 4\left( {DK:x > \dfrac{5}{2}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{9}{2}
\end{array} \right.
\end{array}\)