应《网络安全法》要求，自2017年10月1日起，未进行实名认证将不得使用互联网跟帖服务。为保障您的帐号能够正常使用，请尽快对帐号进行手机号验证，感谢您的理解与支持！

24小时热门版块排行榜    

小木虫论坛-学术科研互动平台 » 计算模拟区 » 计算模拟 » 求解高次复杂函数
北京石油化工学院2026年研究生招生接收调剂公告
121/2返回列表12下一页
查看: 1488  |  回复: 11
只看楼主 @他人 存档 新回复提醒 (忽略) 收藏    在APP中查看

denmarkrico

银虫 (小有名气)

[求助] 求解高次复杂函数 已有3人参与

求解下面这个高次方程时为什么总是得到虚数解，我所用的是MATLAB2012b版

syms h c l theta11 theta21 theta31 x
h=14.5;
l=55.5;
c=19.73;
theta11=0.01/37;
theta21=0.035/37;
theta31=0.065/37;
F=((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) +3^(1/2)*c*sin(theta0) - c*cos(theta0) +l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)))^2 + (((433*c)/250 + (433*h)/250 + 3*l + 2*l*sin(theta11) + 2*l*sin((2*pi)/3 + theta21) - (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) - (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)) - ((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)))^2 + ((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) - c*cos(theta0) + 3^(1/2)*c*sin(theta0) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1)))^2*(2*c*h - 2*l^2*(sin(theta11) - 1) + 2*c^2 - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h)) + ((((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta21) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta21)*(c + h) + 2*c*l*cos(theta0)*cos(theta21) - 2*c*l*sin(theta0)*sin(theta21) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) + ((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)))*(2*c + 2*h + 2*l*cos(theta11) + (2*c*(x^2 - 1))/(x^2 + 1)) - ((3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta21) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta21)*(c + h) + 2*c*l*cos(theta0)*cos(theta21) - 2*c*l*sin(theta0)*sin(theta21) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) - (3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) +l*cos(theta21) - c*cos(theta0) + 3^(1/2)*c*sin(theta0) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)))*(2*l*sin(theta11) - 2*l + (4*c*x)/(x^2 + 1)))*((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) - l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1)))；
回复此楼

» 猜你喜欢

» 本主题相关价值贴推荐，对您同样有帮助:
心 那个凉啊
1楼 2013-12-13 22:28:48
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖
回帖支持 ( 显示支持度最高的前 50 名 )

feixiaolin

荣誉版主 (文坛精英)

优秀版主

【答案】应助回帖

感谢参与，应助指数 +1
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？
赞一下(1人) 回复此楼

» 本帖已获得的红花（最新10朵）

denmarkrico
2楼2013-12-13 22:53:29
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖

ynod23

金虫 (小有名气)

【答案】应助回帖

感谢参与，应助指数 +1
函数没有看完，瞎了。。。
我认为，这种情况，不是查程序哪儿出问题了，而应该看方法对不对。。。。
先试试看能不能把那个大函数写成许多函数相互运算的结果，这样就降低了出错的可能。如果这样还有问题的话，也可以根据matlab的运行提示，找出是哪一行的问题。这样把问题的规模缩小，我觉得比较靠谱。
希望能帮到你。
赞一下(1人) 回复此楼

» 本帖已获得的红花（最新10朵）

denmarkrico
5楼2013-12-14 04:43:11
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖
普通回帖

jerkwin

专家顾问 (正式写手)
你这方程, 能解出来我觉得很奇怪
赞一下 回复此楼
3楼2013-12-13 23:10:14
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖

paopaoai11

木虫 (小有名气)
你这什么方程啊？

[ 发自小木虫客户端 ]
回复此楼
泡泡
4楼2013-12-14 01:03:00
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖

denmarkrico

银虫 (小有名气)
送红花一朵
引用回帖:
2楼: Originally posted by feixiaolin at 2013-12-13 22:53:29
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？

你这个替代法我见过
赞一下 回复此楼
心 那个凉啊
6楼2013-12-14 16:07:30
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖

denmarkrico

银虫 (小有名气)
送红花一朵
引用回帖:
5楼: Originally posted by ynod23 at 2013-12-14 04:43:11
函数没有看完，瞎了。。。
我认为，这种情况，不是查程序哪儿出问题了，而应该看方法对不对。。。。
先试试看能不能把那个大函数写成许多函数相互运算的结果，这样就降低了出错的可能。如果这样还有问题的话，也可 ...

嗯  谢谢你 我这是机器人的正解
赞一下 回复此楼
心 那个凉啊
7楼2013-12-14 16:08:44
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖

denmarkrico

银虫 (小有名气)
引用回帖:
2楼: Originally posted by feixiaolin at 2013-12-13 22:53:29
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？

theta0 是要求的量
赞一下 回复此楼
心 那个凉啊
8楼2013-12-14 16:09:58
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖

denmarkrico

银虫 (小有名气)
引用回帖:
2楼: Originally posted by feixiaolin at 2013-12-13 22:53:29
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？

要解的方程是这样的：

((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - 3^(1/2)*l*sin(theta31) + (3*c*(x^2 - 1))/(x^2 + 1) - (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) - l*sin(theta21) + 3^(1/2)*l*cos(theta21) - (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)))^2*(2*c*h - 2*l^2*(sin(theta11) - 1) + 2*c^2 - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h)) + ((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) - l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - 3^(1/2)*l*sin(theta31) + (3*c*(x^2 - 1))/(x^2 + 1) - (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) - l*sin(theta21) + 3^(1/2)*l*cos(theta21) - (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)))*(((3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - 3^(1/2)*l*sin(theta31) + (3*c*(x^2 - 1))/(x^2 + 1) - (2*3^(1/2)*c*x)/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)) - (3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*(2*l^2*(sin(theta31) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta31)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) + (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)))*(2*l*sin(theta11) - 2*l + (4*c*x)/(x^2 + 1)) - (((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)) + ((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) - l*sin(theta21) + 3^(1/2)*l*cos(theta21) - (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta31) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta31)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) + (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)))*(2*c + 2*h + 2*l*cos(theta11) + (2*c*(x^2 - 1))/(x^2 + 1))) + ((3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - 3^(1/2)*l*sin(theta31) + (3*c*(x^2 - 1))/(x^2 + 1) - (2*3^(1/2)*c*x)/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)) - (3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*(2*l^2*(sin(theta31) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta31)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) + (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)))^2 + (((433*c)/250 + (433*h)/250 + 3*l + 2*l*sin(theta11) + 2*l*sin((2*pi)/3 + theta21) - (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) - (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)) - ((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)))^2;
赞一下 回复此楼
心 那个凉啊
9楼2013-12-14 17:49:07
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖

denmarkrico

银虫 (小有名气)
对不起大家，我用万能公式替代后的方程式一楼的回复所示，此处的x相当于tan(theta0/2)
赞一下 回复此楼
心 那个凉啊
10楼2013-12-14 17:51:45
已阅   回复此楼   关注TA 给TA发消息 送TA红花 TA的回帖
相关版块跳转 我要订阅楼主 denmarkrico 的主题更新
121/2返回列表12下一页
普通表情 龙 兔 虎 猫
最具人气热帖推荐 [查看全部] 作者 回/看 最后发表
[考研] 一志愿南开大学0710生物学359求调剂 +3 兔兔兔111223314 2026-03-29 3/150 2026-03-29 23:14 by Evan_Liu
[考研] 一志愿哈尔滨工业大学材料与化工方向336分 +10 辰沐5211314 2026-03-26 10/500 2026-03-29 21:01 by 唐沐儿
[考研] 070300化学354求调剂 +6 101次希望 2026-03-28 6/300 2026-03-29 12:57 by 无际的草原
[考研] 0856求调剂 +7 楒桉 2026-03-28 7/350 2026-03-29 08:28 by fmesaito
[考研] 295求调剂 +4 wei-5 2026-03-26 4/200 2026-03-28 23:20 by 小木虫tim
[考研] 332求调剂 +4 @MZB382400 2026-03-28 4/200 2026-03-28 21:02 by 唐沐儿
[考研] 320分，材料与化工专业，求调剂 +9 一定上岸aaa 2026-03-27 13/650 2026-03-28 15:00 by 神马都不懂
[考研] 一志愿中南大学化学0703总分337求调剂 +5 niko- 2026-03-27 5/250 2026-03-28 14:25 by 唐沐儿
[考研] 一志愿厦门大学化学学硕307求调剂 +10 y7czhao 2026-03-26 10/500 2026-03-28 14:23 by 唐沐儿
[考研] 085405 考的11408求各位老师带走 +3 Qiu学ing 2026-03-28 3/150 2026-03-28 09:19 by 乐呵呵的追梦人
[考研] 311求调剂 +3 希望上岸阿小杨 2026-03-23 3/150 2026-03-28 07:57 by 热情沙漠
[考研] 材料求调剂 一志愿哈工大总分298分，前三科223分 +5 dongfang59 2026-03-27 5/250 2026-03-28 04:53 by wxiongid
[考研] 315分求调剂 +7 26考研上岸版26 2026-03-26 7/350 2026-03-28 04:05 by fmesaito
[考研] 265求调剂 +8 小木虫085600 2026-03-27 8/400 2026-03-27 22:16 by 无际的草原
[考研] 085600材料与化工调剂 +10 A-哆啦Z梦 2026-03-23 16/800 2026-03-27 15:13 by caszguilin
[考研] 315调剂 +4 0860求调剂 2026-03-26 5/250 2026-03-27 11:23 by wangjy2002
[考研] 292求调剂 +4 求求了收下我吧� 2026-03-26 4/200 2026-03-27 10:37 by zhshch
[考研] 求调剂，一志愿 南京航空航天大学大学 ，080500材料科学与工程学硕 +4 @taotao 2026-03-26 5/250 2026-03-27 08:10 by hypershenger
[考研] 336材料求调剂 +7 陈滢莹 2026-03-26 9/450 2026-03-27 00:20 by wxiongid
[考研] 材料/农业专业，07/08开头均可，过线就行 +3 呵唔哦豁 2026-03-23 4/200 2026-03-23 22:30 by 汪！？！
信息提示
关闭
请填处理意见
关闭