请教一道初二的数学题目。帮帮我。急！！在线等。

先把后边式子分解因式 a²-2a-3\a²-7a+12=（a-3)(a+1)/(a-4)(a-3)
=a+1/a-4 =-1/2

=(a-3)(a+1)/(a-3)(a-4)
=(a+1)/(a-4)
带入
=(2/3+1)/(2/3-4)
=-1/2

二分之三
还是三分之二
（a²-2a-3）\（a²-7a+12）
=（a+1)(a-3)/(a-4)(a-3)=(a+1)/(a-4)=-1
二分之三
三分之二 -1/2

a²-2a-3\a²-7a+12=(a-3)(a+1)\(a-3)(a-4)=(a+1)\(a-4)=(2/3+1)/(2/3-4)
=-1/2

a²-2a-3\a²-7a+12
=（a-3）(a+1)/(a-3)(a-4)
=(a+1)/(a-4)
=(a-4+5)/(a-4)
=1+5/(a-4)
=1+5/(2/3-4)
=1-3/2
=-1/2

用十字相乘法，原始化简为（a+1)(a-3)/(a-3)(a-4),再代入求得得数为-1/2