来自柳林的问题
【解方程1/(x-2)(x-3)-3/(x-1)(x-4)+1/(x-1)(x-2)=1/(x-4)要求用到规律:1/n(n+1)=1/n-1/(n+1)】
解方程1/(x-2)(x-3)-3/(x-1)(x-4)+1/(x-1)(x-2)=1/(x-4)
要求用到规律:1/n(n+1)=1/n-1/(n+1)
1回答
2020-05-29 18:55
【解方程1/(x-2)(x-3)-3/(x-1)(x-4)+1/(x-1)(x-2)=1/(x-4)要求用到规律:1/n(n+1)=1/n-1/(n+1)】
解方程1/(x-2)(x-3)-3/(x-1)(x-4)+1/(x-1)(x-2)=1/(x-4)
要求用到规律:1/n(n+1)=1/n-1/(n+1)
1/(x-2)(x-3)-3/(x-1)(x-4)+1/(x-1)(x-2)=1/(x-4)1/(x-3)-1/(x-2)-1/(x-4)+1/(x-1)+1/(x-2)-1/(x-1)=1/(x-4)1/(x-3)-1/(x-4)=1/(x-4)1/(x-3)=2/(x-4)2(x-3)=x-42x-6=x-4x=2检验是增根所以原分式方程无解...