Solution Let: 3x2−12x+11(x−1)(x−2)(x−3)=Ax−1+Bx−2+Cx−3\frac{3x^2-12x+11}{(x-1)(x-2)(x-3)} =\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}(x−1)(x−2)(x−3)3x2−12x+11=x−1A+x−2B+x−3C Solving gives A=1A=1A=1, B=1B=1B=1, C=1C=1C=1. 3x2−12x+11(x−1)(x−2)(x−3)=1x−1+1x−2+1x−3\boxed{\frac{3x^2-12x+11}{(x-1)(x-2)(x-3)}=\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-3}}(x−1)(x−2)(x−3)3x2−12x+11=x−11+x−21+x−31