(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
请采纳。
解:
1、
1/(x-2)+1/(x-5)=1/(x-3)+1/(x-4)
等号两边的分母常数项分别为2+5=7,3+4=7,解x=7/2
1/(x-7)-1/(x-5)=1/(x-6)-1/(x-4)变形后得1/(x-7)+1/(x-4)=1/(x-6)+1/(x-5)
等号两边的分母常数项分别为7+4=11,6+5=11,解x=7/2
根据以上两个算式得以下规律:
当等号两边的算式均为两项相加,分子为1(或分子为t时),分母为(x-常数),并且等号两边算式得分母常数项和m与n相等时,方程的解x=t*m/2或x=t*n/2
所以变形方程(1/x-a)-(1/x-b)=(1/x-c)-(1/x-d)得,1/(x-a)+1/(x-d)=1/(x-b)+1/(x-c)
根据以上推出的规律的以上方程得解
x=(a+d)/2或x=(b+c)/2
2、(x-1)/(x-2)-(x-3)/(x-4)=(x-2)/(x-3)-(x-4)/(x-5)
变形以上方程得
(x-1)/(x-2)+(x-4)/(x-5)=(x-2)/(x-3)+(x-3)/(x-4)
化简方程过程为以下:
(x-2+1)/(x-2)+(x-5+1)/(x-5)=(x-3+1)/(x-3)+(x-4+1)/(x-4)
(x-2)/(x-2)+1/(x-2)+(x-5)/(x-5)+1/(x-5)=(x-3)/(x-3)+1/(x-3)+(x-4)/(x-4)+1/(x-4)
1+1/(x-2)+1+1/(x-5)=1+1/(x-3)+1+1/(x-4)
1/(x-2)+1/(x-5)=1/(x-3)+1/(x-4)
根据第一个问题得出的规律
方程得解
x=(2+5)/2=7/2
(1)X=(a+d)/2=(b+c)/2
(2)X=7/2