Question

解方程：x-7分之1-x-5分之1=x-6分之1-x-4分之1 能告诉下方法么？

Answer 1

1/(x-7)-1/(x-5)=1/(x-6)-1/(x-4)
1/(x-6)-1/(x-7)=1/(x-4)-1/(x-5)
(x-7)/(x-6)(x-7)-(x-6)/(x-6)(x-7)=(x-5)/(x-4)(x-5)-(x-4)/(x-4)(x-5)
[(x-7)-(x-6)]/(x-6)(x-7)=[(x-5)-(x-4)]/(x-4)(x-5)
(x-7-x+6)/(x-6)(x-7)=(x-5-x+4)/(x-4)(x-5)
-1/(x-6)(x-7)=-1/(x-4)(x-5)
1/(x-6)(x-7)=1/(x-4)(x-5)
(x-6)(x-7)=(x-4)(x-5)
x^2-13x+42=x^2-9x+20
13x-9x=42-20
4x=22
x=11/2
经检验x=11/2是方程有解

Answer 2

x-7分之1-x-5分之1=x-6分之1-x-4分之1
(x-5-x+7)/(x-7)(x-5)=(x-4-x+6)/(x-6)(x-4)
2/(x²-12x+35)=2/(x²-10x+24)
交叉相乘得：
x²-10x+24=x²-12x+35
2x=11
x=11/2

Answer 3

等式两边通分 可得
2/[（x-7）(x-5)]=2/[(x-6)(x-4)]
也就是（x-7）(x-5)=(x-6)(x-4)
x的方-12x+35=x的方-10x+24
2x=11
x=5.5

Answer 4

1/(x-7)-1/(x-5)=1/(x-6)-1/(x-4)
[x-5-x+7)/(x-7)(x-5)=(x-4-x+6)/(x-6)(x-4)
(x-6)(x-4)=(x-7)(x-5)
x^2-10x+24=x^2-12x+35
2x=11
x=5.5

Answer 5

通分
左边分子分母都*（x-7）（x-5）
右边分子分母都*（x-6）（x-4）
比较麻烦啊