首先,题目的后项是不是少了个括号?是[(x-1)/2y-1]?是1/(2y).还是 (1/2)y[(x-1/2y)²-1]-[(x-1)/2y-1]=(x-1)²/4y²-1-[(x-1)/2y-1]=[(x-1)²-4y²]/4y²-[(x-1)/2y-1]=(x-1+2y)(x-1-2y)/4y&...对不起。。。[(x-1/2y)²-1]-[x-1/2y-1]²是这个式子���ȣ����е����Ϊʲô������Ҫ�������š����������Ű�=[(x-1/2y)²-1²]-[x-1/2y-1]²=(x-1/2y-1)(x-1/2y+1)-[x-1/2y-1]²=(x-1/2y-1)[(x-1/2y+1-([x-1/2y-1)]=(x-1/2y-1)(x-1/2y+1-x+1/2y+1)=2(x-1/2y-1)=2x-y-2=[(x-1/2y)²-1²]-[x-1/2y-1]²=(x-1/2y-1)(x-1/2y+1)-[x-1/2y-1]²=(x-1/2y-1)[(x-1/2y+1-([x-1/2y-1)]后面的 平方呢=(x-1/2y-1)(x-1/2y+1-x+1/2y+1)=2(x-1/2y-1)=2x-y-2��ȡ����ʽ(x-1/2y-1)��Ȼ����������������ʽ���������ˣ��������ڵ���һ����