Question

(x-4/x-5)-(x-7/x-8)=(x-5/x-6)-(x-8/x-9)解方程 帮我解出来，写过程，我没学过一元二次方程。

Answer 1

两边分别通分： 1/(x-5)(x-8)=1/(x-6)(x-9)然后两边都取倒数.. (x-5)(x-8)=(x-6)(x-9)展开得 2x=14所以x=7

Answer 2

容易！-3/（x-5)(x-8)=-3/(x-6)(x-9) 3/x的平方-13x+40=3/x的平方-15x+45 x的平方-13x+40=x的平方-15x+45 x=5/2

Answer 3

x-4/x-5-x+7/x+8=x-5/x-6-x+8/x+9-4/x+7/x+3=-5/x+8/x+33/x+3=3/x+3结果为0，你这好像不是一元二次方程

Answer 4

(x-4/x-5)-(x-7/x-8)=(x-5/x-6)-(x-8/x-9)(x-4)/(x-5)+(x-8)/(x-9)=(x-7)/(x-8)+(x-5)/(x-6)
1+1/(x-5)+1+1/(x-9)=1+1/(x-8)+1+1/(x-6)
1/(x-5)+1/(x-9)=1/(x-8)+1/(x-6)
1/(x-5)-1/(x-8)=1/(x-6)-1/(x-9)
-3/(x^2-13x+40)=-3/(x^2-15x+54)
x^2-13x+40=x^2-15x+54
2x=14
x=7