- 一道初二数学题(x+2)/(x+1)+(x+6)/(X+5)=(
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(x+2)/(x+1)+(x+6)/(X+5)=(x+3)/(x+2)+)=(x+5)/(x+4)
- 解:原题:(x+2)/(x+1)+(x+6)/(x+5)=(x+3)/(x+2)+(x+5)/(x+4)
因为(x+2)/(x+1)=(x+1+1)/(x+1)=1+1/(x+1)
(x+6)/(x+5)=(x+5+1)/(x+5)=1+1/(x+5)
(x+3)/(x+2)=(x+2+1)/(x+2)=1+1/(x+2)
(x+5)/(x+4)=(x+4+1)/(x+4)=1+1/(x+4)
原方程化为:1/(x+1)+1/(x+5)=1/(x+2)+1/(x+4)
移项:1/(x+5)-1/(x+2)=1/(x+4)-1/(x+1)
两边分别通分,(x+5-x-2)/(x+5)(x+2)=(x+4-x-1)/(x+4)(x+1)
整理,3/(x+5)(x+2)=3/(x+4)(x+1)
变形,(x+5)(x+2)=(x+4)(x+1)
解这个方程,得x=-3