分式的计算

(2/x^2-x-2 )-(3/x^2+5x+4)
={2/[(x-2)(x+1)]}-{3/[(x+1)(x+4)]}
=[1/(X+1)]*{[2/(X-2)]-[3/(X+4)]}
=[1/(X+1)]*{(2X+8-3X+6)/[(X-2)(X+4)]}
=(14-X)/[(X+1)(X-2)(X+4)]

请问^是什么意思

首先，去括号

得2/x^2-x-2 -3/x^2-5x-4

交换，得2/x^2-3/x^2-x-5x-2-4
=（-1）/x^2-6x-6

(2/x²-x-2 )-(3/x²+5x+4)
=2/x²-x-2-3/x²-5x-4
=-5X-6-1/x²