计算题！！！！！！！！！！！！！
两边各减2,即(x+1)/(x-2)-1 + (x-1)/(x-3)-1 = (x-3)/(x-4)-1 + (x+3)/(x-1)-1
可化为 (x+1-x+2)/(x-2) + (x-1-x+3)/(x-3) = (x-3-x+4)/(x-4) + (x+3-x+1)/(x-1)
再化简得3/(x-2) + 2/(x-3) = 1/(x-4) + 4/(x-1)
两边的式子变换一下得 3/(x-2) - 4/(x-1) = 1/(x-4) - 2/(x-3)
再通分得3(x-1)-4(x-2)/(x-2)(x-1) = (x-3)-2(x-4)/(x-3)(x-4)
再化简得(x-1)...全部
两边各减2,即(x+1)/(x-2)-1 + (x-1)/(x-3)-1 = (x-3)/(x-4)-1 + (x+3)/(x-1)-1
可化为 (x+1-x+2)/(x-2) + (x-1-x+3)/(x-3) = (x-3-x+4)/(x-4) + (x+3-x+1)/(x-1)
再化简得3/(x-2) + 2/(x-3) = 1/(x-4) + 4/(x-1)
两边的式子变换一下得 3/(x-2) - 4/(x-1) = 1/(x-4) - 2/(x-3)
再通分得3(x-1)-4(x-2)/(x-2)(x-1) = (x-3)-2(x-4)/(x-3)(x-4)
再化简得(x-1)(x-2) = (x-3)(x-4)
再展开得x^2-3x+2 = x^2-7x+12
再化简得4x=10
最后解得x=2。
5
好麻烦,打了好长时间,如果看得懂就把分给我吧,不懂再问我。收起
