宇宙战将武器作弊码:化简:x(x+1)\1+(x+1)(x+2)\1+(x+2)(x+3)\1+(x+3)(x+4)\1

来源:百度文库 编辑:神马品牌网 时间:2024/04/28 05:20:15
应该系化简:化简:x(x+1)/1+(x+1)(x+2)/1+(x+2)(x+3)/1+(x+3)(x+4)/1

如果1是分子,就应该这么算
原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/x-1/(x+4)
如果1是分母,这题就没意思了,把括号都打开,合并同类项即可

1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/x-1/(x+4)=3/(x(x-4))

x(x+1)\1+(x+1)(x+2)\1+(x+2)(x+3)\1+(x+3)(x+4)\1
=x\1-(x+1)\1+.........(x+3)\1-(x+4)\1
=x\1-(x+4)\1
=x(x+4)\4

/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/x-1/(x+4)=3/(x(x-4))

x(x+1)\1+(x+1)(x+2)\1+(x+2)(x+3)\1+(x+3)(x+4)\1
=x\1-(x+1)\1+.........(x+3)\1-(x+4)\1
=x\1-(x+4)\1
=x(x+4)\4