\(a.\left|x\right|=\dfrac{2}{3}\\ =>x=\pm\dfrac{2}{3}\\ b.\left|x\right|=-\dfrac{4}{-7}=\dfrac{4}{7}\\ =>x=\pm\dfrac{4}{7}\\ c.\left|x\right|=\left|-\dfrac{3}{2}\right|=\dfrac{3}{2}\\ =>x=\pm\dfrac{3}{2}\\ d.\left|x\right|=-\dfrac{5}{-2}=\dfrac{5}{2}\\ =>x=\pm\dfrac{5}{2}\\ e.\left|x\right|=5\dfrac{1}{2}=\dfrac{11}{2}\\ =>x=\pm\dfrac{11}{2}\\ f.\left|x\right|=\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{-1}{20}< 0\\ =>x\in\varnothing\)
g: \(\left|x+1\right|=\dfrac{1}{4}\)
=>\(\left[{}\begin{matrix}x+1=\dfrac{1}{4}\\x+1=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}-1=-\dfrac{3}{4}\\x=-\dfrac{1}{4}-1=-\dfrac{5}{4}\end{matrix}\right.\)
h: \(\left|x-1\right|=\dfrac{2}{3}\)
=>\(\left[{}\begin{matrix}x-1=\dfrac{2}{3}\\x-1=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1+\dfrac{2}{3}=\dfrac{5}{3}\\x=1-\dfrac{2}{3}=\dfrac{1}{3}\end{matrix}\right.\)
i: \(\left|x+2\right|=\dfrac{1}{12}\)
=>\(\left[{}\begin{matrix}x+2=\dfrac{1}{12}\\x+2=-\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{12}-2=-\dfrac{23}{12}\\x=-\dfrac{1}{12}-2=-\dfrac{25}{12}\end{matrix}\right.\)
j: \(\left|x-4\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-4=\dfrac{1}{2}\\x-4=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4+\dfrac{1}{2}=\dfrac{9}{2}\\x=4-\dfrac{1}{2}=\dfrac{7}{2}\end{matrix}\right.\)
k: \(\left|x+3\right|=\dfrac{4}{5}\)
=>\(\left[{}\begin{matrix}x+3=\dfrac{4}{5}\\x+3=-\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}-3=\dfrac{4}{5}-\dfrac{15}{5}=-\dfrac{11}{5}\\x=-\dfrac{4}{5}-3=-\dfrac{4}{5}-\dfrac{15}{5}=-\dfrac{19}{5}\end{matrix}\right.\)
i: \(\left|x-7\right|=-\dfrac{5}{3}\)
mà \(\left|x-7\right|>=0\forall x\)
nên \(x\in\varnothing\)
m: \(\left|x-7\right|=\dfrac{5}{3}\)
=>\(\left[{}\begin{matrix}x-7=\dfrac{5}{3}\\x-7=-\dfrac{5}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7+\dfrac{5}{3}=\dfrac{26}{3}\\x=7-\dfrac{5}{3}=\dfrac{21}{3}-\dfrac{5}{3}=\dfrac{16}{3}\end{matrix}\right.\)
n: \(\left|x-\dfrac{1}{2}\right|=\dfrac{1}{4}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{4}\\x-\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}\\x=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\end{matrix}\right.\)
o: \(\left|x-\dfrac{3}{2}\right|=1\dfrac{1}{3}\)
=>\(\left|x-\dfrac{3}{2}\right|=\dfrac{4}{3}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{4}{3}\\x-\dfrac{3}{2}=-\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}+\dfrac{3}{2}=\dfrac{8}{6}+\dfrac{9}{6}=\dfrac{17}{6}\\x=-\dfrac{4}{3}+\dfrac{3}{2}=-\dfrac{8}{6}+\dfrac{9}{6}=\dfrac{1}{6}\end{matrix}\right.\)
p: \(\left|x-\dfrac{5}{4}\right|=-\dfrac{1}{3}\)
mà \(\left|x-\dfrac{5}{4}\right|>=0\forall x\)
nên \(x\in\varnothing\)