Question

初二解分式方程

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

Answer 1

解：(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)
首先 等号两边分子相同
且 x^2-11+28不等于x^2-11+30
分母不相等，要等式成立只能分子为0
所以 2x-11=0
解得 x=11/2
经检验，x=11/2是原方程的解
希望有帮助 谢谢了

Answer 2

(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 3

(x-5)/(x-7)+(x-2)/(x-4)=(x-3)/(x-5)+(x-4)/(x-6) ，
移项得：
(x-5)/(x-7)-(x-4)/(x-6)=(x-3)/(x-5)-(x-2)/(x-4)
2/(x-7）(x-6)=2/(x-5)(x-4)
(x-7）(x-6)=(x-5)(x-4)
-13x+42=-9x+20
4x=22
x=5.5

Answer 4

是x-7吗？
先化简的到1/(x-7)+1/(x-4)=1/(x-5)+1/(x-6);
接着化简就可得到2x-11=0；
x=5.5