ta có:
\(\left|2x+\frac{1}{7}\right|=\left|-2x-\frac{1}{7}\right|;\left|-2x-\frac{1}{7}\right|\ge-2x-\frac{1}{7}\)
\(\left|2x+\frac{1}{6}\right|\ge0;\left|2x+\frac{1}{5}\right|\ge2x+\frac{1}{5}\)
=> \( A\ge2x+\frac{1}{5}+0-2x-\frac{1}{7}=\frac{2}{35}\)
dấu "=" xảy ra <=>\(x=-\frac{1}{12}\)