A Bx+c 例3 (1+2x)(1+x2)1+2x1+x2 l=A(1+x)+(Bx+C(l+2x), 整理得1=(A+2B)x2+(B+2C)x+C+A, A+2B=0, B+2C=0,→ 2-525 5 A+C=1 x 5十 (1+2x)(1+x2)1+2x1+x2例3 . 1 5 1 5 2 1 2 5 4 2 x x x + − + + + = (1 2 )(1 ) 1 2 + x + x 1 (1 ) ( )(1 2 ), 2 = A + x + Bx + C + x 1 ( 2 ) ( 2 ) , 2 = A+ B x + B + C x + C + A      + = + = + = 1, 2 0, 2 0, A C B C A B , 5 1 , 5 2 , 5 4  A = B = − C = , 1 2 1 2 x Bx C x A + + + + = (1 2 )(1 ) 1 2 + x + x  整理得