Ta có : \(a-b=7\Rightarrow a=b+7\) (1)
Thay (1) vào \(\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}\) ta có:
\(\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}=\frac{3\left(b+7\right)-b}{2\left(b+7\right)+7}+\frac{3b-b-7}{2b-7}\)
\(=\frac{3b+21-b}{2b+14+7}+\frac{2b-7}{2b-7}\)
\(=\frac{2b+21}{2b+21}+1\)
\(=1+1\)
\(=2\)
Vậy \(\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}=2\)