\(B=8x+\dfrac{6}{x}+18y+\dfrac{7}{y}=\left(8x+\dfrac{2}{x}\right)+\left(18y+\dfrac{2}{y}\right)+\left(\dfrac{4}{x}+\dfrac{5}{y}\right)\ge8+12+23=43\)
Dấu bằng xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{3}\right)\)
Vậy, \(MinB\) là \(43\) khi \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{3}\right)\)