Question

数学，求详解！！！

x+1分之x-x+2分之x+1-x+3分之x+2+x+4分之x+3=
x+1分之x+2-x+2分之x+3-x-3分之x-4+x-4分之x-5=
5分钟内急求！！！回答详细提高悬赏！！！

Answer 1

x+1分之x-x+2分之x+1-x+3分之x+2+x+4分之x+3
=x+x-x+x/2+1-x+x/3+2+x+x/4+3
=（1+2+3）+（x+x/2+x/3+x/4）
=6+2又12分之1x

x+1分之x+2-x+2分之x+3-x-3分之x-4+x-4分之x-5
=x+x+2-x+x/2+3-x-x/3-4+x-x/4-5
=（2+3-4-5）+(x+x/2-x/3-x/4)
=-4+12分之11x

刚看到，不算晚吧。觉得有帮助，给个采纳吧。

Answer 2

x+1分之x-x+2分之x+1-x+3分之x+2+x+4分之x+3
=x/(x+1)-(x+1)/(x+2)-(x+2)/(x+3)+(x+3)/(x+4)
=(x²+2x-x²-2x-1)/(x+1)(x+2)+(x²+6x+9-x²-6x-8)/(x+3)(x+4)
=-1/(x+1)(x+2)+1/(x+3)(x+4)
=(-x²-7x-12+x²+3x+2)/(x+1)(x+2)(x+3)(x+4)
=(-4x-10)/[(x+1)(x+2)(x+3)(x+4)]

x+1分之x+2-x+2分之x+3-x-3分之x-4+x-4分之x-5
=(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
=(x²+4x+4-x²-4x-3)/(x+1)(x+2)-(x²-8x+16-x²+8x-15)/(x-3)(x-4)
=1/(x+1)(x+2)-1/(x-3)(x-4)
=(x²-7x+12-x²+3x+2)/(x+1)(x+2)(x-3)(x-4)
=(-4x+14)/[(x+1)(x+2)(x-3)(x-4)]