Question

B=3/4.b+4/3.b-1/2.b với b=6/19

Answer 1

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 \(B=\dfrac{1}{2}\)