Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} d.\ x\ \in \left\{1;\frac{-7}{6}\right\}\\ e.\ x\ \in \left\{\frac{116}{35} ;\frac{-4}{65}\right\}\\ f.\ x\ \in \left\{\frac{55}{28} ;\frac{-45}{52}\right\} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} d.\ |7x+6|-\mid 5x+8|=0\\ \Leftrightarrow |7x+6|=\mid 5x+8|\\ TH1:\ 7x+6=5x+8\\ \Leftrightarrow 2x=2\\ \Leftrightarrow x=1\\ TH2:7x+6=-( 5x+8)\\ \Leftrightarrow 12x=-14\\ \Leftrightarrow x=-\frac{7}{6}\\ e.\ \mid \frac{5}{4} x-\frac{3}{2} \mid -\mid \frac{3}{8} x+\frac{7}{5} \mid =0\\ \Leftrightarrow \mid \frac{5}{4} x-\frac{3}{2} \mid =\mid \frac{3}{8} x+\frac{7}{5} \mid \\ TH1:\ \frac{5}{4} x-\frac{3}{2} =\frac{3}{8} x+\frac{7}{5}\\ \Leftrightarrow \frac{7}{8} x=\frac{29}{10}\\ \Leftrightarrow x=\frac{116}{35}\\ TH2:\frac{5}{4} x-\frac{3}{2} =-\left(\frac{3}{8} x+\frac{7}{5}\right)\\ \Leftrightarrow \frac{13}{8} x=\frac{1}{10}\\ \Leftrightarrow x=\frac{}{}\frac{4}{65}\\ f.\ \mid \frac{2}{5} x+\frac{5}{3} \mid -\mid \frac{4}{3} x-\frac{1}{6} \mid =0\\ \Leftrightarrow \mid \frac{2}{5} x+\frac{5}{3} \mid =\mid \frac{4}{3} x-\frac{1}{6} \mid \\ TH1:\ \frac{2}{5} x+\frac{5}{3} =\frac{4}{3} x-\frac{1}{6}\\ \Leftrightarrow \frac{-14}{15} x=\frac{-11}{6}\\ \Leftrightarrow x=\frac{55}{28}\\ TH2:\ \frac{2}{5} x+\frac{5}{3} =-\left(\frac{4}{3} x-\frac{1}{6}\right)\\ \Leftrightarrow \frac{26}{15} x=\frac{-3}{2}\\ \Leftrightarrow x=\frac{-45}{52}\\ \end{array}$
