阿Q吧 > 解方程(x-1⼀x-2)+(x-7⼀x-8)=(x-3⼀x-4)+(x-5⼀x-6)

解方程(x-1⼀x-2)+(x-7⼀x-8)=(x-3⼀x-4)+(x-5⼀x-6)

谢谢，要过程

2024-11-18 03:25:53

推荐回答（1个）

回答1：

(x-1)/(x-2)+(x-7)/(x-8)=(x-3)/(x-4)+(x-5)/(x-6)
(x-2+1)/(x-2)+(x-8+1)/(x-8)=(x-4+1)/(x-4)+(x-6+1)/(x-6)
(此步对分子加1减1，是简便算法)
1+1/(x-2)+1+1/(x-8)=1+1/(x-4)+1+1/坦好(x-6)
1/(x-2)+1/(x-8)=1/(x-4)+1/(x-6)
1/(x-8)-1/局弯(x-4)=1/(x-6)-1/(x-2)
(此步移项成左右相减，是桐信闷再次简便算法)
(x-4-x+8)/(x-8)(x-4)=(x-2-x+6)/(x-6)(x-2)
4/(x-8)(x-4)=4/(x-6)(x-2)
(x-8)(x-4)=(x-6)(x-2)
x^2-12x+32=x^2-8x+12
4x=20
x=5
验算:
1/3-1/3=1-1(验算也正确)