初二数学 解方程 x+2⼀x+1+x+8⼀x+7=x+6⼀x+5+x+4⼀x+3

首先由题意得x+1≠0，x+7≠0，x+5≠0，x+3≠0，
即x≠-1，x≠-7，x≠-5，x≠-3，则先简化方程
(x+1+1)/(x+1)+(x+7+1)/(x+7)=(x+5+1)/(x+5)+(x+3+1)/(x+3)

即1+1/(x+1)+1+1/(x+7)=1+1/(x+5)+1+1/(x+3)
1/(x+1)+1/(x+7)=1/(x+5)+1/(x+3)

1/(x+1)-1/(x+3)=1/(x+5)-1/(x+7) 两边通分可得

[(x+3)-(x+1)]/[(x+3)(x+1)]=[(x+7)-(x+5)]/[(x+7)(x+5)]

即2/[(x+3)(x+1)]=2/[(x+7)(x+5)]
∴(x+3)(x+1)=(x+7)(x+5)
x²+4x+3=x²+12x+35
4x+3=12x+35

x=-4
经检验，x=-4是原方程的解！

(x+2)/(X+1)-1+(x+8)/(x+7)-1=(x+6)/(x+5)-1+(x+4)/(x+3)-1
1/(x+1)+1/(x+7)=1/(x+5)+1/(x+3)
1/(x+1)-1/(x+3)=1/(x+5)-1/(x+7)
[(x+3)-(x+1)]/[(x+1)(x+3)]=[(x+7)-(x+5)]/[(x+5)(x+7)]
2/(x^2+4x+3)=2/(x^2+12x+35)
x^2+4x+3=x^2+12x+35
4x+3=12x+35
x=-4