用Mathematica解方程组,输出如下错误,何意?另考虑到是否为无解,请问用该软怎么判断方程组是否有解?输入Solve[(y1 - y2)/(x1 - x2) = (55 - 16)/(-8 - 13),(x1 - x2)^2 + (y1 - y2)^2 = (55 - 16)^2 + (-8 - 13)^2,(x2 - 5)^2 +

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/05 11:52:14
用Mathematica解方程组,输出如下错误,何意?另考虑到是否为无解,请问用该软怎么判断方程组是否有解?输入Solve[(y1 - y2)/(x1 - x2) = (55 - 16)/(-8 - 13),(x1 - x2)^2 + (y1 - y2)^2 = (55 - 16)^2 + (-8 - 13)^2,(x2 - 5)^2 +
xU]oH+[I}"i ږ? Q)b[h@h E-P4-BMY >/읙8 UsνsqɨK/IyJ )l+?ͅl ߽x1x[$=G?)'IFq -UzÍAtmDFԮ \Hħ:fAӥ -Ct!JTqg~2DL; 3Yl)yP8)`2||^MuQ8lbBjzU;gJIː Ѱ

用Mathematica解方程组,输出如下错误,何意?另考虑到是否为无解,请问用该软怎么判断方程组是否有解?输入Solve[(y1 - y2)/(x1 - x2) = (55 - 16)/(-8 - 13),(x1 - x2)^2 + (y1 - y2)^2 = (55 - 16)^2 + (-8 - 13)^2,(x2 - 5)^2 +
用Mathematica解方程组,输出如下错误,何意?另考虑到是否为无解,请问用该软怎么判断方程组是否有解?
输入Solve[(y1 - y2)/(x1 - x2) = (55 - 16)/(-8 -
13),(x1 - x2)^2 + (y1 - y2)^2 = (55 - 16)^2 + (-8 -
13)^2,(x2 - 5)^2 + (y2 - 10)^2 = (13 - 5)^2 + (16 -
10)^2,(x1 - 5)^2 + (y1 - 10)^2 = (-8 - 5)^2 + (55 - 10)^2,{x1,
x2,y1,y2}]
输出Set::write:Tag Times in (y1-y2)/(x1-x2) is Protected.>>
Set::write:Tag Plus in (x1-x2)^2+(y1-y2)^2 is Protected.>>
Set::write:Tag Plus in (-5+x2)^2+(-10+y2)^2 is Protected.>>
General::stop:Further output of Set::write will be suppressed during this calculation.>>
Solve::nonopt:Options expected (instead of {x1,x2,y1,y2}) beyond position 4 in Solve[-(13/7),1962,100,2194,{x1,x2,y1,y2}].An option must be a rule or a list of rules.>>
上面的输入出错误了,少了一个“=”。
输入改正为下:Solve[(y1 - y2)/(x1 - x2) == (55 - 16)/(-8 -
13),(x1 - x2)^2 + (y1 - y2)^2 == (55 - 16)^2 + (-8 -
13)^2,(x2 - 5)^2 + (y2 - 10)^2 == (13 - 5)^2 + (16 -
10)^2,(x1 - 5)^2 + (y1 - 10)^2 == (-8 - 5)^2 + (55 -
10)^2,{x1,x2,y1,y2}]
输出为:Solve::nonopt:Options expected (instead of {x1,x2,y1,y2}) beyond position 4 in Solve[(y1-y2)/(x1-x2)==-(13/7),(x1-x2)^2+(y1-y2)^2==1962,(-5+x2)^2+(-10+y2)^2==100,(-5+x1)^2+(-10+y1)^2==2194,{x1,x2,y1,y2}].An option must be a rule or a list of rules.>>
输出提示为何意?用何种方法判断方程组是否有解?急...

用Mathematica解方程组,输出如下错误,何意?另考虑到是否为无解,请问用该软怎么判断方程组是否有解?输入Solve[(y1 - y2)/(x1 - x2) = (55 - 16)/(-8 - 13),(x1 - x2)^2 + (y1 - y2)^2 = (55 - 16)^2 + (-8 - 13)^2,(x2 - 5)^2 +
所有的关系方程必须用{ }括起来,正确的为:
Solve[{(y1 - y2)/(x1 - x2) == (55 - 16)/(-8 -
13),(x1 - x2)^2 + (y1 - y2)^2 == (55 - 16)^2 + (-8 -
13)^2,(x2 - 5)^2 + (y2 - 10)^2 == (13 - 5)^2 + (16 -
10)^2,(x1 - 5)^2 + (y1 - 10)^2 == (-8 - 5)^2 + (55 -
10)^2},{x1,x2,y1,y2}]
解出来有两组解,运行结果为:
{{x1 -> 18,x2 -> -3,y1 -> -35,y2 -> 4},{x1 -> 3860/109,x2 -> 1571/109,y1 -> (2793/109),
y2 -> 1458/109},
{x1 -> -(2770/109),x2 -> -(481/109),y1 -> 4973/109,y2 -> 722/109},{x1 -> -8,x2 -> 13,y1 -> 55,y2 -> 16}}
可以给积分了吧!

关于Mathematica解方程组的问题,输出时没有错误,解不出来的原因如题,我在一个三元一次方程组时,应该能解出来的,但是老是得不到结果.由于我刚刚接触Mathematica这个软件,不懂,所以我把解得编 用mathematica解含积分的微分方程就是这个题.mathematica解不出来啊,直接原样输出了 mathematica解方程组,结果还是原式子. 用mathematica软件,求解三角函数方程组.形如:Solve[{Cos (x y) == 1,Sin (x y) == 0},{x,y}].为什么mathematica软件计算无解.(其实解为xy=Pi/2) 怎么用Mathematica解方程组的实数解,而复数解省略如题,在用Mathematica解方程组时,方程组可能有实数解和复数解,但是我只关心实数解,即我只要实数解,复数解省略,同时将解方程组的实数解运用 Mathematica中,如何在函数中输出图像 mathematica中如何在定义的函数中输出图形 有关mathematica用Dsolve解偏微分方程组的格式如图,想解出偏微分方程的x,y,求Dsolve命令格式 mathematica教程如何用mathematica解方程组?是Slove[(ax+by==0,cx+dy==0),(x,y)] 怎么将Mathematica解方程组的解应用到后续的运算中啊?如题,怎么将Mathematica上一个方程组的解应用到后续的运算中.例如用Mathematica已经得到三元一次方程组2a+b+3c=13,2b-a=3,3b-c=3的解a=1,b=2,c=3,怎么 mathematica解三角方程请问如何用mathematica解这个方程? 用Mathematica解方程组,输出如下错误,何意?另考虑到是否为无解,请问用该软怎么判断方程组是否有解?输入Solve[(y1 - y2)/(x1 - x2) = (55 - 16)/(-8 - 13),(x1 - x2)^2 + (y1 - y2)^2 = (55 - 16)^2 + (-8 - 13)^2,(x2 - 5)^2 + 变系数非线性常微分方程组用mathematica怎么求解? 怎么用mathematica求解含参微分方程组 如何用 mathematica 解常微分方程? 怎么用mathematica把一个函数的定义域输入,输出值域,和输入值域,输出定义域? Mathematica如何方程的解自动输出等号形式?Mathematica解方程时如何自动输出等号形式?例如方程:Reduce[x + 2 a == 10,{x}],输出x == 10 - 2 a,是双等号形式.我需要单等号形式:x = 10 - 2 a,怎么得到?用Tradit mathematica中如何按实际比例输出