\(49x^2-22x+9=\left(7x\right)^2-2.7.\dfrac{11}{7}x+\dfrac{121}{49}+\dfrac{320}{49}\)
\(=\left(7x-\dfrac{11}{7}\right)^2+\dfrac{320}{49}\ge\dfrac{320}{49}\) dấu"=" xảy ra<=>\(x=\dfrac{11}{49}\)
\(=>\sqrt{49x^2-22x+9}\ge\)\(\sqrt{\dfrac{320}{49}}=\dfrac{8\sqrt{5}}{7}\)
\(=>B\ge\dfrac{8\sqrt{5}}{7}+8\sqrt{38}\)