If $\frac{2x-5}{{\left(x-3\right)}^2}=\frac{a}{\left(x-3\right)}+\frac{b}{{\left(x-3\right)}^2}$ then a+b=

$\frac{2x-5}{{\left(x-3\right)}^2}=\frac{a}{\left(x-3\right)}+\frac{b}{{\left(x-3\right)}^2} ----\left(1\right)$

$\frac{2x-5}{{\left(x-3\right)}^2}=\frac{a(x-3)+b}{{\left(x-3\right)}^2}$

by canceling (x-3)²       (common both side)

$2x-5=a\left(x-3\right)+b$

$2x-5=x·a+(-3a+b)$

on comparing we will get 

a=2

and -3a+b=-5

putting this value of a

the:

-3$\times$2+b=-5

-6+b=-5

b=-5+6

b=1

then a+b=2+1

a+b=3

the optiob C is correct.