Xác định các số hữu tỷ a,b,c sao cho: (9x^2-16x+4)/(x^3-3x^2+2x)=a/x+b/(x-1)+c/(x-2)

=>9x2−16x+4x(x−1)(x−2)=a(x−1)(x−2)+bx(x−2)+cx(x−1)x(x−1)(x−2)=>9x2−16x+4x(x−1)(x−2)=x2(a+b+c)+(−3a−2b−c)x+2ax(x−1)(x−2)=>a+b+c=9−3a−2b−c=−162a=4=>a=2b=3c=4.