a,\(\left|2x+1\right|=\frac{3}{5}\)
\(2x+1=+-\frac{3}{5}\)
\(\left[\begin{array}{nghiempt}2x+1=\frac{3}{5}\Rightarrow x=-\frac{1}{5}\\2x+1=-\frac{3}{5}\Rightarrow x=-\frac{4}{5}\end{array}\right.\)
Vậy \(x\in\left\{-\frac{1}{5};-\frac{4}{5}\right\}\)
\(\frac{x+3}{4}=-\frac{7}{2}\)
\(\Leftrightarrow2\left(x+3\right)=4.\left(-7\right)\)
\(2\left(x+3\right)=-28\)
\(\left(x+3\right)=28:2\)
\(x+3=14\\\)
\(\Rightarrow x=14-3\)
\(x=11\)
\(\left(\frac{7}{2}-2x\right).\frac{4}{3}=\frac{22}{3}\)
\(\left(\frac{7}{2}-2x\right)=\frac{22}{3}:\frac{4}{3}\)
\(\left(\frac{7}{2}-2x\right)=\frac{11}{2}\)
\(2x=\frac{7}{2}-\frac{11}{2}\)
\(2x=2\)
\(x=1\)
\(\left(\frac{7}{3}+\frac{7}{2}\right).x=-\frac{25}{6}+\frac{22}{7}\)
\(\frac{35}{6}.x=-\frac{43}{42}\)
\(x=-\frac{43}{42}:\frac{35}{6}\)
\(x=-\frac{43}{245}\)