初二数学题

显示全部楼层 · 2013-3-17 17:36:28

根据题意，对上式进行变形得：(x-4)/(x-5)=(x-5+1)/(x-5)=1+1/(x-5)同理：（x-8)/(x-9)=1+1/(x-9)（x-7)/(x-8)=1+1/(x-8)（x-5)/(x-6)=1+1/(x-6)所以方程变形得：1+1/(x-5) +1+1/(x-9)=1+1/(x-8)+1+1/(x-6)化简得：1/(x-5) +1/(x-9)=1/(x-8)+1/(x-6)移项得：1/(x-5)-1/(x-6)= 1/(x-8)-1/(x-9) 等式两边分别通分得：（x-6-x+5)/(x-6)(x-5)=(x-9-x+8)/(x-8)(x-9)(x-6)(x-5)=(x-8)(...

千问 · 2013-3-17 17:36:28

原方程即1+1/(x-5)+1+1/(x-9)=1+1/(x-8)+1+1/(x-6)1/（x-5)+1/(x-9)=1/(x-8)+1/(x-6)1/(x-5)-1/(x-6)=1/(x-8)-1/(x-9)-1/(x-5)(x-6)=-1/(x-8)(x-9)(x-5)(x-6)=(x-8）（x-9)x2-11x+30=...