Theo bài ra, ta có:
\(a-b=3\Rightarrow a=b+3\)
Thay \(a=b+3\) vào \(B\), ta có:
\(B=\dfrac{a-8}{b-5}-\dfrac{4a-b}{3a+3}\\ B=\dfrac{b+3-8}{b-5}-\dfrac{4\left(b+3\right)-b}{3\left(b+3\right)+3}\\ B=\dfrac{b+3-8}{b-5}-\dfrac{4\left(b+3\right)-b}{3\left(b+3\right)+3}\\ B=\dfrac{b-5}{b-5}-\dfrac{4b+12-b}{3b+9+3}\\ B=1-\dfrac{3b+12}{3b+12}\\ B=1-1\\ B=0\)
Vậy: \(B=0\)
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Chúc bạn học tốt 
theo bài ra ta có:
\(B=\frac{a-8}{b-5}-\frac{4a-b}{3a+3}\)
\(\Rightarrow B=\frac{a-8}{b-5}-1-\frac{4a-b}{3a+3}+1\)
\(\Rightarrow B=\left(\frac{a-8}{b-5}-1\right)+\left(1-\frac{4a-b}{3a+3}\right)\)
\(\Rightarrow B=\frac{a-8-\left(b-5\right)}{b-5}+\frac{3a+3-\left(4a-b\right)}{3a+3}\)
\(\Rightarrow B=\frac{a-8-b+5}{b-5}+\frac{3a+3-4a+b}{3a+3}\)
\(\Rightarrow B=\frac{a-b-8+5}{b-5}+\frac{b-a+3}{3a+3}\) \(\Rightarrow B=\frac{3-3}{b-5}+\frac{-3+3}{3a+3}\)
\(\Rightarrow B=0+0\\ \Rightarrow B=0\)
vậy B = 0
