(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)
答案:3 悬赏:40 手机版
解决时间 2021-02-27 23:35
- 提问者网友：低吟詩仙的傷
- 2021-02-27 19:40
(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)
最佳答案
- 五星知识达人网友：西风乍起
- 2021-02-27 20:51
原式=1/（x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/(x+1)-1/(x+2)+1/(x+2)-1/（x+3)+1/(x+3)-1/(x+4)
=1/(x+1)-1/(x+4)
=(x+4-x-1)/(x²+5x+4)
=3/(x²+5x+4)
祝你开心
=1/(x+1)-1/(x+2)+1/(x+2)-1/（x+3)+1/(x+3)-1/(x+4)
=1/(x+1)-1/(x+4)
=(x+4-x-1)/(x²+5x+4)
=3/(x²+5x+4)
祝你开心
全部回答
- 1楼网友：不如潦草
- 2021-02-27 23:20
(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)
=1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/(x+1)-1/(x+4)
=3/[(x+1)(x+4)]
=1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/(x+1)-1/(x+4)
=3/[(x+1)(x+4)]
- 2楼网友：雪起风沙痕
- 2021-02-27 21:50
写错了吧，估计应该是1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)
1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)
=1/((x+1)(x+2))+1/((x+2)(x+3))+1/((x+3)(x+4))
=((x+3)(x+4)+(x+1)(x+4)+(x+1)(x+2))/((x+1)(x+2)(x+3)(x+4))
=(3x^2+15x+18)/((x+1)(x+2)(x+3)(x+4))
=(3(x+2)(x+3))/((x+1)(x+2)(x+3)(x+4))
=3/((x+1)(x+4))
=3/(x^2+5x+4)
1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)
=1/((x+1)(x+2))+1/((x+2)(x+3))+1/((x+3)(x+4))
=((x+3)(x+4)+(x+1)(x+4)+(x+1)(x+2))/((x+1)(x+2)(x+3)(x+4))
=(3x^2+15x+18)/((x+1)(x+2)(x+3)(x+4))
=(3(x+2)(x+3))/((x+1)(x+2)(x+3)(x+4))
=3/((x+1)(x+4))
=3/(x^2+5x+4)
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
大家都在看
推荐资讯