Question

如何解3x-3/x-2+x-5/x-8=3x-9/x-4+x-3/x-6

Answer 1

解：
(3x-3)/(x-2)+(x-5)/(x-8)=(3x-9)/(x-4)+(x-3)/(x-6)
(3x-3)/(x-2)-(3x-9)/(x-4)=(x-3)/(x-6)-(x-5)/(x-8)
3[(x-1)/(x-2) -(x-3)/(x-4）]=(x-3)/(x-6)-(x-5)/(x-8)
左右同时通分得
3[(x-1)(x-4)-(x-3)(x-2)]/(x-2)(x-4)=[(x-3)(x-8)-(x-5)(x-6)]/(x-6)(x-8)
3（x²-5x+4-x²+5x-6)/(x-2)(x-4) =(x²-11x+24-x²+11x-30)/(x-6)(x-8)
3×（-2）/(x-2)(x-4)=（-6）/(x-6)(x-8)
1/(x-2)(x-4)=1/(x-6)(x-8)
(x-2)(x-4)=(x-6)(x-8)
x²-6x+8=x²-14x+48
-6x+14x=48-8
8x=40
x=5

Answer 2

(3x-6+3)/(x-2)+(x-8+3)/(x-8)=(3x-12+3)/(x-4)+(x-6+3)/(x-6)
3+3/(x-2)+1+3/(x-8)=3+3/(x-4)+1+3/(x-6) 两边减去4
3/(x-2)+3/(x-8)=3/(x-4)+3/(x-6) 两边除以3
1/(x-2)+1/(x-8)=1/(x-4)+1/(x-6)
1/(x-2)-1/(x-4)=1/(x-6)-1/(x-8)
(x-4-x+2)/(x-2)(x-4)=(x-8-x+6)/(x-8)(x-6)
-2/(x-2)(x-4)=-2/(x-8)(x-6)
(x-8)(x-6)=(x-2)(x-4)
x²-14x+48=x²-6x+8
-8x=-40
x=5

Answer 3

方程的两边都乘以x2，消去分母上的x，然后求解，最后验证x不等于0，因为分母不能等于0