Đặt 761=a; 139=b
\(M=\left(3+\dfrac{1}{3b}\right)\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\left(4+\dfrac{a}{a+1}\right)-\dfrac{4}{\left(a+1\right)\cdot3b}+\dfrac{5}{b}\)
\(=\dfrac{9b+1}{3b}\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\dfrac{4a+4}{a+1}-\dfrac{4}{3b\left(a+1\right)}+\dfrac{5}{b}\)
\(=\dfrac{9b+1-3\left(4a+4\right)-4+15\left(a+1\right)}{3b\left(a+1\right)}\)
\(=\dfrac{9b+1-12a-12-4+15a+15}{3b\left(a+1\right)}\)
\(=\dfrac{3a+9b}{3b\left(a+1\right)}=\dfrac{3\left(a+3b\right)}{3b\left(a+1\right)}=\dfrac{a+3b}{b\left(a+1\right)}\)
\(=\dfrac{761+3\cdot139}{139\left(761+1\right)}=\dfrac{589}{52959}\)