Question

解方程:(x-5)/(x-7)+(x-2)/(x-4)=(x-3)/(x-5)+(x-4)/(x-6)

Answer 1

(x-5)/(x-7)+(x-2)/(x-4)=(x-3)/(x-5)+(x-4)/(x-6)
(x-7+2)/(x-7)+(x-4+2)/(x-4)=(x-5+2)/(x-5)+(x-6+2)/(x-6)
1-2/(x-7)+1-2/(x-4)=1-2/(x-5)+1-2/(x-6)
2/(x-7)+2/(x-4)=2/(x-5)+2/(x-6)
1/(x-7)+1/(x-4)=1/(x-5)+1/(x-6)
(x-4+x-7)/(x-4)(x-7)=(x-6+x-5)/(x-5)(x-6)
(2x-11)/(x^2-11+28)=(2x-11)/(x^2-11+30)
因为x^2-11+28不等于x^2-11+30
分母不等，要等式成立只能分子为0
2x-11=0
x=11/2
经检验，x=11/2是原方程的解

Answer 2

题出错了还是你再忽悠人呢~
X X平方都能消去~最后43=49~

Answer 3

X=11/2

Answer 4

首先等式两边同时乘以（x-3）*(x-5)的因式，去分母原等式等价于 （x-6）*(x-5)+(x-4)*（x-3）=2（x-3）*(x-5)展开 x^2-11x+30+x^2-7x+12=2x^2-16x+30移项，合并同类项 2x=12x=6
x=3

x=5
时等式无意义。所以x=6
可取。