(6x-17)/(x^2-5x+6)=A/(x-3)+B/(x-2).
(Ax-2A+Bx-3B)/(x^2-5x+6). Matching terms we get A+B=6 and 2A+3B=17. Double the first equation: 2A+2B=12 and subtract this from the second equation: B=5, therefore A=6-5=1. The partial fractions are 1/(x-3)+5/(x-2).