Ta có:\(\frac{3a-b}{2a+15}=\frac{3a-b}{2a+a-b}=\frac{3a-b}{3a-b}=1\)
\(\frac{3b-a}{2b-15}=\frac{3b-a}{2b-\left(a-b\right)}=\frac{3b-a}{3b-a}=1\)
=>P=1+1=2
Ta có a = 15 + b
=> \(\frac{3a-b}{2a+15}+\frac{3b-a}{2b-15}\) = \(\frac{3\left(15+b\right)-b}{2\left(15+b\right)+15}+\frac{3b-\left(15+b\right)}{2b-15}\)
= \(\frac{45+3b-b}{30+2b+15}+\frac{3b-15-b}{2b-15}\)
= \(\frac{45+2b}{45+2b}+\frac{2b-15}{2b-15}\)= 1 + 1 = 2