你看到如1/[(y-4)(y+1)]这类的分式可以先拆成1/(y-4) - 1/(y+1)通分之后加减得到5/[(y-4)(y+1)]即1/(y-4) - 1/(y+1)=5/[(y-4)(y+1)]则1/5[1/(y-4) - 1/(y+
你看到如1/[(y-4)(y+1)]这类的分式
可以先拆成1/(y-4) - 1/(y+1)
通分之后加减得到5/[(y-4)(y+1)]
即1/(y-4) - 1/(y+1)=5/[(y-4)(y+1)]
则1/5[1/(y-4) - 1/(y+1)]=1/[(y-4)(y+1)]
此类裂项所有分母相乘的分式都适用,但是要注意未知量x(或y,n)前面的系数两个括号内都要一样才能裂项,如:
1/[(n-1)(n+1)]
先裂成1/(n-1) - 1/(n+1) = 2/[(n-1)(n+1)]
则1/2[1/(n-1) - 1/(n+1)] = 1/[(n-1)(n+1)]
若:
1/[(2n-1)(n+1)]则无法裂项
