→Ax(x+1)+B(x+1)+Cx2=1 A+C=0 →〈A+B=0→{B=1 B=1 C A B 但若2 x2(x+1)x2x+1 →A(x+1)+Bx2=1 →A=0,=1矛盾( 1) ( 1) 1 2  Ax x + + B x + + Cx =  = + = + =  1 00 BA B A C  === −  111 CBA 但若 ( 1) 1 1 2 2 + = + + x B xA x x( 1) 1 2  A x + + Bx =  A = 0,A = 1 矛盾