本文介绍了如何求解系统方程式x ^ 2-4x + y ^ 2-2y + 1 = 0;的x ^ 2-8x + Y ^ 2-2y + 8 = 0?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想在C#中解决这个方程组: (x-2)^ 2 +(y-1)^ 2 = 2 ^ 2 (x-4)^ 2 +(y-1)^ 2 = 3 ^ 2 等于: x ^ 2 -4x + y ^ 2 -2y +1 = 0 x ^ 2 -8x + y ^ 2 -2y +8 = 0 但是,我不知道.Net Framework是否有解决方案或我必须使用外部库 你能帮助我吗? 感谢所有
I want to solve this system of equations in C #: (x-2) ^ 2 + (y-1) ^ 2 = 2 ^ 2 (x-4) ^ 2 + (y-1) ^ 2 = 3 ^ 2 is equal to: x^2 -4x +y^2 -2y +1 = 0 x^2 -8x +y^2 -2y +8 = 0 But, I don´t know if .Net Framework have a solution or I must use a external library Can you help me somebody? Thanks for all
推荐答案你可能想看看微软解决方案基金会 [ ^ ]可以为您提供... You may want to see what Microsoft Solver Foundation[^] can provide you...
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