解分式方程1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)1/(x

1个回答

  • 1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)+1/(x²+9x+20)=1/8

    1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=1/8

    1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)=1/8

    1/(x+1)-1/(x+5)=1/8

    8(x+5)-8(x+1)=(x+1)(x+5)

    8x+40-8x-8=x²+6x+5

    x²+6x-27=0

    (x+9)(x-3)=0

    x=-9或x=3

    经检验x=-9或x=3都是方程的根