一道初二数学题(x+2)/(x+1)+(x+6)/(X+5)=(
(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