dễ thấy |x+2/7| > 0;|x+4/7|>0;|x+3 1/7| >0
=>|x+2/7|+|x+4/7|+|x+3 1/7| > 0;mà VT=VP
nên 4x>0
ta có: \(\left|x+\frac{2}{7}\right|+\left|x+\frac{4}{7}\right|+\left|x+3\frac{1}{7}\right|=4x=>x+\frac{2}{7}+x+\frac{4}{7}+x+\frac{22}{7}=4x=>3x+4=4x=>x=4\)
vậy x=4