Đáp án:

 $1.1\\
a) {\left[\begin{aligned}x=\dfrac{9}{2}\\x=\dfrac{1}{2}\end{aligned}\right.}\\
b) {\left[\begin{aligned}x=\dfrac{7}{12}\\x=\dfrac{4}{6}=\dfrac{2}{3}\end{aligned}\right.}\\
c) {\left[\begin{aligned} x=\dfrac{-1}{30}\\x =\dfrac{-11}{30}\end{aligned}\right.}\\$
d) Vô nghiệm
$1.2\\
a){\left[\begin{aligned} x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{aligned}\right.}\\
b) 
 {\left[\begin{aligned} x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{aligned}\right.}\\
c)
 {\left[\begin{aligned} x=\dfrac{20}{15}=\dfrac{4}{3}\\x=\dfrac{-28}{15}\end{aligned}\right.}\\
1.3\\
a)
{\left[\begin{aligned} x=1\\x=\dfrac{-1}{3}\end{aligned}\right.}\\
b)
 {\left[\begin{aligned} x=8\\x=-4\end{aligned}\right.}\\
c)
 {\left[\begin{aligned} x=\dfrac{-13}{5}\\x=\dfrac{17}{5}\end{aligned}\right.}\\
d)
{\left[\begin{aligned} x=\dfrac{38}{15}\\x=\dfrac{-28}{15}\end{aligned}\right.}\\$

Giải thích các bước giải:

 $1.1\\
a) |2x-5|=4\\
\Leftrightarrow {\left[\begin{aligned}2x-5=4\\2x-5=-4\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=9\\2x=1\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{9}{2}\\x=\dfrac{1}{2}\end{aligned}\right.}\\
b) \dfrac{1}{3}-\left | \dfrac{5}{4}-2x \right |=\dfrac{1}{4}\\
\Leftrightarrow\left | \dfrac{5}{4}-2x \right |=\dfrac{1}{3}-\dfrac{1}{4} \\
\Leftrightarrow\left | \dfrac{5}{4}-2x \right |=\dfrac{4}{12}-\dfrac{3}{12} \\
\Leftrightarrow\left | \dfrac{5}{4}-2x \right |=\dfrac{1}{12} \\
\Leftrightarrow {\left[\begin{aligned}\dfrac{5}{4}-2x=\dfrac{1}{12}\\\dfrac{5}{4}-2x =\dfrac{-1}{12}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=\dfrac{5}{4}-\dfrac{1}{12}\\2x=\dfrac{5}{4}+\dfrac{1}{12}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=\dfrac{15}{12}-\dfrac{1}{12}\\2x=\dfrac{15}{12}+\dfrac{1}{12}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=\dfrac{14}{12}=\dfrac{7}{6}\\2x=\dfrac{16}{12}=\dfrac{4}{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{7}{12}\\x=\dfrac{4}{6}=\dfrac{2}{3}\end{aligned}\right.}\\
c) \dfrac{1}{2}-\left | x+\dfrac{1}{5} \right |=\dfrac{1}{3}\\
\Leftrightarrow \left | x+\dfrac{1}{5} \right |=\dfrac{1}{2}-\dfrac{1}{3}\\
\Leftrightarrow \left | x+\dfrac{1}{5} \right |=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\\
\Leftrightarrow {\left[\begin{aligned}x+\dfrac{1}{5} =\dfrac{1}{6}\\x+\dfrac{1}{5} =\dfrac{-1}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{1}{6}-\dfrac{1}{5}\\x =\dfrac{-1}{6}-\dfrac{1}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{5}{30}-\dfrac{6}{30}\\x =\dfrac{-5}{30}-\dfrac{6}{30}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{-1}{30}\\x =\dfrac{-11}{30}\end{aligned}\right.}\\
d) \dfrac{3}{4}-\left | 2x+1 \right |=\dfrac{7}{8}\\
\Leftrightarrow |2x+1|=\dfrac{3}{4}-\dfrac{7}{8}\\
\Leftrightarrow |2x+1|=\dfrac{6}{8}-\dfrac{7}{8}\\
\Leftrightarrow |2x+1|=\dfrac{-1}{8}(VN)\\
1.2\\
a)2|2x-3|=\dfrac{1}{2}\\
\Leftrightarrow |2x-3|=\dfrac{1}{4}\\
\Leftrightarrow {\left[\begin{aligned} 2x-3=\dfrac{1}{4}\\2x-3=\dfrac{-1}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 2x=\dfrac{1}{4}+3\\2x=\dfrac{-1}{4}+3\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 2x=\dfrac{1}{4}+\dfrac{12}{4}\\2x=\dfrac{-1}{4}+\dfrac{12}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 2x=\dfrac{13}{4}\\2x=\dfrac{11}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{aligned}\right.}\\
b) 
7,5-3|5-2x|=-4,5\\
\Leftrightarrow 3|5-2x|=7,5+4,5=12\\
\Leftrightarrow |5-2x|=4\\
\Leftrightarrow  {\left[\begin{aligned} 5-2x=4\\5-2x=-4\end{aligned}\right.}\\
\Leftrightarrow  {\left[\begin{aligned} 2x=1\\2x=9\end{aligned}\right.}\\
\Leftrightarrow  {\left[\begin{aligned} x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{aligned}\right.}\\
c)
\left | x+\dfrac{4}{15} \right |-|-3,75|=-|-2,15|\\
\Leftrightarrow \left | x+\dfrac{4}{15} \right |-3,75=-2,15\\
\Leftrightarrow \left | x+\dfrac{4}{15} \right |=-2,15+3,75=\dfrac{8}{5}\\
\Leftrightarrow  {\left[\begin{aligned} x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{aligned}\right.}\\
\Leftrightarrow  {\left[\begin{aligned} x=\dfrac{8}{5}-\dfrac{4}{15}\\x=\dfrac{-8}{5}-\dfrac{4}{15}\end{aligned}\right.}\\
\Leftrightarrow  {\left[\begin{aligned} x=\dfrac{24}{15}-\dfrac{4}{15}\\x=\dfrac{-24}{15}-\dfrac{4}{15}\end{aligned}\right.}\\
\Leftrightarrow  {\left[\begin{aligned} x=\dfrac{20}{15}=\dfrac{4}{3}\\x=\dfrac{-28}{15}\end{aligned}\right.}\\
1.3\\
a)
2|3x-1|+1=5\\
\Leftrightarrow 2|3x-1|=4\\
\Leftrightarrow |3x-1|=2\\
\Leftrightarrow {\left[\begin{aligned} 3x-1=2\\3x-1=-2\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 3x=3\\3x=-1\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=1\\x=\dfrac{-1}{3}\end{aligned}\right.}\\
b)
\left | \dfrac{x}{2}-1 \right |=3\\
\Leftrightarrow {\left[\begin{aligned} \dfrac{x}{2}-1=3\\\dfrac{x}{2}-1=-3\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} \dfrac{x}{2}=4\\\dfrac{x}{2}=-2\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=8\\x=-4\end{aligned}\right.}\\
c)
\left | -x+\dfrac{2}{5} \right |+\dfrac{1}{2}=3,5\\
\Leftrightarrow \left | -x+\dfrac{2}{5} \right |=\dfrac{7}{2}-\dfrac{1}{2}=3\\
\Leftrightarrow {\left[\begin{aligned} -x+\dfrac{2}{5}=3\\-x+\dfrac{2}{5}=-3\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{2}{5}-3\\x=\dfrac{2}{5}+3\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{2}{5}-\dfrac{15}{5}\\x=\dfrac{2}{5}+\dfrac{15}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{-13}{5}\\x=\dfrac{17}{5}\end{aligned}\right.}\\
d)
\left | x-\dfrac{1}{3} \right |=2\dfrac{1}{5}=\dfrac{11}{5}\\
\Leftrightarrow {\left[\begin{aligned} x-\dfrac{1}{3}=\dfrac{11}{5}\\x-\dfrac{1}{3}=\dfrac{-11}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{11}{5}+\dfrac{1}{3}\\x=\dfrac{-11}{5}+\dfrac{1}{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{33}{15}+\dfrac{5}{15}\\x=\dfrac{-33}{15}+\dfrac{5}{15}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{38}{15}\\x=\dfrac{-28}{15}\end{aligned}\right.}\\$