1/2-1/3+|x|=x+1/6 |1/5-2x|.|1/3-4x|=0 |x|+2x-1/3=-5/6 |x-1/3|+(x-1/4)+1/2=x -3/7-|3/7+x|=1/10

+, $\frac{1}{2}$ – $\frac{1}{3}$ + |x| = x + $\frac{1}{6}$

⇔ $\frac{1}{6}$ – |x| = x + $\frac{1}{6}$

⇔ |x| = -x (vô lí)

Vì |x| ≥ 0

⇒ Phương trình vô nghiệm

+, |$\frac{1}{5}$ – 2x| . |$\frac{1}{3}$ – 4x| = 0

⇔ \(\left[ \begin{array}{l}|\frac{1}{5}-2x|=0\\|\frac{1}{3}-4x|=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{1}{5}-2x=0\\\frac{1}{3}-4x=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}2x=\frac{1}{5}\\4x=\frac{1}{3}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{1}{10}\\x=\frac{1}{12}\end{array} \right.\) 

Vậy x = $\frac{1}{10}$ hoặc $\frac{1}{12}$

+, |x| + 2x – $\frac{1}{3}$ = -$\frac{5}{6}$

⇔ |x| = -$\frac{1}{2}$ – 2x

⇔  \(\left[ \begin{array}{l}x=-\frac{1}{2}-2x\\x=\frac{1}{2}+2x\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}3x=-\frac{1}{2}\\-x=\frac{1}{2}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-\frac{1}{6}\\x=-\frac{1}{2}\end{array} \right.\) 

Vậy ………

+, |x – $\frac{1}{3}$| + (x – $\frac{1}{4}$) + $\frac{1}{2}$ = x

⇔ |x – $\frac{1}{3}$| = -$\frac{1}{4}$

⇔ \(\left[ \begin{array}{l}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{7}{12}\\x=\frac{1}{12}\end{array} \right.\) 

Vậy ……….

+, -$\frac{3}{7}$ – |$\frac{3}{7}$ + x| = $\frac{1}{10}$

⇔ |$\frac{3}{7}$ + x| = -$\frac{37}{70}$

Vì |$\frac{3}{7}$ + x| ≥ 0 ⇒ |$\frac{3}{7}$ + x| = -$\frac{37}{70}$ (vô lí)

⇒ Phương trình vô nghiệm