4, | $\frac{3}{2}$x + $\frac{1}{2}$ = |4x - 1|
⇔ \(\left[ \begin{array}{l}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\frac{3}{2}x-4x=-1-\frac{1}{2}\\\frac{3}{2}x+4x=1-\frac{1}{2}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{3}{5}\\x=\frac{1}{11}\end{array} \right.\)
5, |$\frac{7}{5}$x + $\frac{1}{2}$| = |$\frac{4}{3}$x - $\frac{1}{4}$|
⇔ \(\left[ \begin{array}{l}\frac{7}{5}x+\frac{1}{2}=\frac{4}{3}x-\frac{1}{4}\\\frac{7}{5}x+\frac{1}{2}=\frac{1}{4}-\frac{4}{3}x\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\frac{7}{5}x-\frac{4}{3}x=-\frac{1}{4}-\frac{1}{2}\\\frac{7}{5}x+\frac{4}{3}x=\frac{1}{4}-\frac{1}{2}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\frac{1}{15}x=-\frac{3}{4}\\\frac{41}{15}=-\frac{1}{4}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-\frac{45}{4}\\x=-\frac{15}{164}\end{array} \right.\)
6, (4x - 9)(2,5 + $\frac{-7}{3}$x) = 0
⇔ \(\left[ \begin{array}{l}4x-9=0\\2,5-\frac{7}{3}x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}4x=9\\\frac{7}{3}x=2,5\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{9}{4}\\x=\frac{15}{14}\end{array} \right.\)
7, $\frac{3}{2}$ - (x - $\frac{5}{6}$) = $\frac{8}{9}$
⇔ x - $\frac{5}{6}$ = $\frac{11}{18}$
⇔ x = $\frac{13}{9}$
