原式=(a²+3)/[(a+1)(a-1)]-(a+1)/(a-1)+1
=[(a²+3)-(a+1)²+a²-1]/[(a+1)(a-1)]
=[a²+3-a²-2a-1+a²-1]/[(a+1)(a-1)]
=(a²-2a+1)/[(a+1)(a-1)]
=(a-1)²/[(a+1)(a-1)]
=(a-1)/(a+1)
第一题 你做得对
原式=(a²+3)/[(a+1)(a-1)]-(a+1)/(a-1)+1
=[(a²+3)-(a+1)²+a²-1]/[(a+1)(a-1)]
=[a²+3-a²-2a-1+a²-1]/[(a+1)(a-1)]
=(a²-2a+1)/[(a+1)(a-1)]
=(a-1)²/[(a+1)(a-1)]
=(a-1)/(a+1)
第一题 你做得对