搜搜做题作业网mathematica解符号变量方程
来源：学生作业帮 编辑：搜搜做题作业网作业帮 分类：综合作业 时间：2024/04/28 00:26:00
mathematica解符号变量方程
A + B + C == 0,
A (b + c) + B (a + c) + C (a + b) == 0,
Abc + Bac + Cab == 1
三个方程联立要解出A,B,C来,而且A,B,C的符号表达式完全由a,b,c表示.
A + B + C == 0,
A (b + c) + B (a + c) + C (a + b) == 0,
Abc + Bac + Cab == 1
三个方程联立要解出A,B,C来,而且A,B,C的符号表达式完全由a,b,c表示.
In[23]:= Solve[{A + B + C1 == 0,
A (b + c) + B (a + c) + C1 (a + b) == 0,
A b c + B a c + C1 a b == 1},{A,B,C1}]
{{A -> -(1/((-a + b) (a - c))),B -> -(1/((a - b) (b - c))),
C1 -> -(1/((a - c) (-b + c)))}}
记得加空格,否则连写则认为是一个变量名.
A (b + c) + B (a + c) + C1 (a + b) == 0,
A b c + B a c + C1 a b == 1},{A,B,C1}]
{{A -> -(1/((-a + b) (a - c))),B -> -(1/((a - b) (b - c))),
C1 -> -(1/((a - c) (-b + c)))}}
记得加空格,否则连写则认为是一个变量名.