你是不是x-4少打了个括号?
1/(x-2)(x-3)+1/(x-3)(x-4)-1/(x-4)=1
如果是:逐步通分
左=1/(x-2)(x-3)+1/(x-3)(x-4)-(x-3)/(x-3)(x-4)
=1/(x-2)(x-3)+(1-x+3)/(x-3)(x-4)
=1/(x-2)(x-3)-1/(x-3)
=(1-x+2)/(x-2)(x-3)
=-1/(x-2)=1
所以x=1
如果我理解错了请追问。
这道题是用1/(x-2)(x-3)=[1/(x-3)]-[1/(x-2)]这样的形式来做的吧,不需要把分母乘上去
最后分解得到[1/(x-3)]-[1/(x-2)]+[1/(x-4)]-[1/(x-3)]-[1/(x-4)]=1
即-[1/(x-2)]=1
x=1
[1/(x-3)]-[1/(x-2)]+[1/(x-4)]-[1/(x-3)]-[1/(x-4)]=1
即-[1/(x-2)]=1
x=1
3x²+8x+3/x²+4x+11=3x²+8x+9/x²+4x+13为神马算出来无解呢……