Môn Toán Lớp 7: |x- 1/5|=|-5/2|-(1/4 – 2/3) |x+5/6|=|1/5-2/3|+ -3/4 |x- 1/5|+1/3=1/4-|-3/2|

\[\begin{array}{l}
\# andy\\
a)\left| {x – \dfrac{1}{5}} \right| = \left| { – \dfrac{5}{2}} \right| – \left( {\dfrac{1}{4} – \dfrac{2}{3}} \right)\\
 \Rightarrow \left| {x – \dfrac{1}{5}} \right| = \dfrac{5}{2} – \left( { – \dfrac{5}{{12}}} \right)\\
 \Rightarrow \left| {x – \dfrac{1}{5}} \right| = \dfrac{5}{2} + \dfrac{5}{{12}}\\
 \Rightarrow \left| {x – \dfrac{1}{5}} \right| = \dfrac{{35}}{{12}}\\
 \Rightarrow \left[ \begin{array}{l}
x – \dfrac{1}{5} = \dfrac{{35}}{{12}}\\
x – \dfrac{1}{5} =  – \dfrac{{35}}{{12}}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{187}}{{60}}\\
x =  – \dfrac{{163}}{{60}}
\end{array} \right.\\
 \Rightarrow S \in \left\{ {\dfrac{{187}}{{60}}; – \dfrac{{163}}{{60}}} \right\}\\
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\
b)\left| {x + \dfrac{5}{6}} \right| = \left| {\dfrac{1}{5} – \dfrac{2}{3}} \right| + \left( { – \dfrac{3}{4}} \right)\\
 \Rightarrow \left| {x + \dfrac{5}{6}} \right| = \left| { – \dfrac{7}{{15}}} \right| – \dfrac{3}{4}\\
 \Rightarrow \left| {x + \dfrac{5}{6}} \right| = \dfrac{7}{{15}} – \dfrac{3}{4}\\
 \Rightarrow \left| {x + \dfrac{5}{6}} \right| =  – \dfrac{{17}}{{60}}\left( {Vô\,\,lý} \right)\\
 \Rightarrow S \in \left\{ \emptyset  \right\}\\
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\
c)\left| {x – \dfrac{1}{5}} \right| + \dfrac{1}{3} = \dfrac{1}{4} – \left| { – \dfrac{3}{2}} \right|\\
 \Rightarrow \left| {x – \dfrac{1}{5}} \right| + \dfrac{1}{3} = \dfrac{1}{4} – \dfrac{3}{2}\\
 \Rightarrow \left| {x – \dfrac{1}{5}} \right| + \dfrac{1}{3} =  – \dfrac{5}{4}\\
 \Rightarrow \left| {x – \dfrac{1}{5}} \right| =  – \dfrac{5}{4} – \dfrac{1}{3}\\
 \Rightarrow \left| {x – \dfrac{1}{5}} \right| =  – \dfrac{{19}}{{12}}\left( {Vô\,\,lý} \right)\\
 \Rightarrow S \in \left\{ \emptyset  \right\}
\end{array}\]