Đặt \(a=\frac{1}{33}\), \(b=\frac{1}{59}\)
Có B= \(\left(2+\frac{1}{33}\right).\frac{1}{59}-3.\frac{1}{33}.\left(3+\frac{58}{59}\right)-4.\frac{1}{33}\frac{1}{59}+4.\frac{1}{33}.3\)
= \(\left(2+a\right)b-3a\left(3+1-\frac{1}{59}\right)-4ab+4.a.3\)
= \(2b+ab-3a\left(4-b\right)-4ab+12a\)
= \(2b+ab-12a+3ab-4ab+12a\)
= \(2b=\frac{2}{59}\)
Vậy B= \(\frac{2}{59}\)