\(A=\sqrt{\left(7x-3\right)^2}+\sqrt{\left(7x+3\right)^2}\)
\(A=\left|7x-3\right|+\left|7x+3\right|=\left|3-7x\right|+\left|7x+3\right|\)
\(A\ge\left|3-7x+7x+3\right|=6\)
\(A_{min}=6\) khi \(\left(3-7x\right)\left(7x+3\right)\ge0\Rightarrow-\frac{3}{7}\le x\le\frac{3}{7}\)