题目

解方程：（x+1)/(x+2)+（x+6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6),为什么必须使方程两边分母常数差相等?
我的解法：原方程=
（x+1)/(x+2)+(x+6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6)
1-1/(x+2)+1-1/(x+7)=1-1/(x+3)+1-1/(x+6)
-1/(x+2)-1/(x+7)=-1/(x+3)-1/(x+6)
1/(x+2)+1/(x+7)=1/(x+3)+1/(x+6)
(2x+9)/(x^2+9x+14)=(2x+9)/(x^2+9x+18)
为什么这样就求不出来,只能按
1/(x+2)-1/(x+3)=1/(x+6)-1/(x+7)
(x+3-(x+2))/(x+2)(x+3)=(x+7-(x+6))/(x+6)(x+7)
1/(x+2)(x+3)=1/(x+6)(x+7)
(x+2)(x+3)=(x+6)(x+7)
x^2+5x+6=x^2+13x+42
8x=-36
x=-9/2
这样做?按道理应该也行的.
知其然而要知其所以然.
风唱梵音

提问时间：2020-06-13

答案

一样可以解出来.(2x+9)/(x^2+9x+14)=(2x+9)/(x^2+9x+18)移项提取公因式：(2x+9)[1/(x^2+9x+14)-1/(x^2+9x+18)]=0则有2x+9=0 =>x=-9/2或：1/(x^2+9x+14)-1/(x^2+9x+18)=0,通分后得：4/【(x^2+9x+14)*(x^2+9x+18)】=0...