Ta có: \(B=\dfrac{3}{4}\cdot b+\dfrac{4}{3}\cdot b-\dfrac{1}{2}\cdot b\)
\(=b\left(\dfrac{3}{4}+\dfrac{4}{3}-\dfrac{1}{2}\right)\)
\(=b\left(\dfrac{9}{12}+\dfrac{16}{12}-\dfrac{6}{12}\right)\)
\(=\dfrac{19}{12}\cdot b\)
Thay \(b=\dfrac{6}{19}\) vào B, ta được:
\(B=\dfrac{6}{19}\cdot\dfrac{19}{12}=\dfrac{6}{12}=\dfrac{1}{2}\)
Vậy: Khi \(b=\dfrac{6}{19}\) thì \(B=\dfrac{1}{2}\)