方程(1/x-2)+(1/x-5)=(1/x-3)+(1/x-4)的解是x=7/2.方程(1/x-7)-(1/x-5)=

3个回答

  • (1)x=(a+d)/2

    第一个方程中方程两边的分母中常数项的和都是7,x的值是7/2,而(1/x-a)-(1/x-b)=(1/x-c)-(1/x-d)可变形为:(1/x-a)+(1/x-d)=(1/x-c)+(1/x-b),其中a+d=b+c,所以x=(a+d)/2

    (2)x=7/2

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

    移项得:(x-1)/(x-2)+(x-4)(x-5)=(x-2)/(x-3)+(x-3)/(x-4)

    变形得:-(x-1)/(x-2)-(x-4)(x-5)=-(x-2)/(x-3)-(x-3)/(x-4)

    方程两边都加2:1-(x-1)/(x-2)+1-(x-4)(x-5)=1-(x-2)/(x-3)+1-(x-3)/(x-4)

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

    x=7/2