Question

因式分解 帮忙解一下

Answer 1

原题=(x²+3x+9)/(x³-27)+6x/(9-x²)-(x-1)/(6+2x)??
=(x²+3x+3²)/(x³-3³)-6x/[(x+3)(x-3)]-(x-1)/[2(x+3)]
=(x²+3x+3²)/[(x-3)(x²+3x+3²)]-12x/[2(x+3)(x-3)]-(x-1)(x-3)/[2(x+3)(x-3)]
=2(x+3)/[2(x+3)(x-3)]-12x/[2(x+3)(x-3)]-(x-1)(x-3)/[2(x+3)(x-3)]
=(2x+6-12x-x²+3x+x-3)/[2(x+3)(x-3)]
=(-x²-6x+3)/[2(x+3)(x-3)]

追问

都是x的2次方的

追答

(x²+3x+9)/(x²-27)+6x/(9x-x²)-(x-1)/(6+2x)
=(x²+3x+9)/(x²-27)+6x/[x(9-x)]-(x-1)/[2(3+x)]
=(x²+3x+9)/(x²-27)+6*2(x+3)/[2(x+3)(9-x)]-(x-1)(9-x)/[2(3+x)(9-x)]
=(x²+3x+9)/(x²-27)+(12x+36-9x+x²+9-x)/[2(27+6x-x²)]
=(x²+3x+9)/(x²-27)+(x²+2x+45)/[2(27+6x-x²)]
=(x²+3x+9)/(x²-27)-(x²+2x+45)/[(x+3)(x-9)]
...
请仔细核对下原题（也可能有印刷错误），太复杂了。