作业帮matlab二元函数请问各路大神,MatLab中我求二元二次函数,怎么会这样的结果,看不懂啊,>> [xc,yc]=so
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/17 09:09:15
matlab二元函数
请问各路大神,MatLab中我求二元二次函数,怎么会这样的结果,看不懂啊,
>> [xc,yc]=solve('2*(xa-xc)^2+2*(ya-yc)^2=(xb-xa)^2+(yb-ya)^2','2*(xb-xc)^2+2*(yb-yc)^2=(xb-xa)^2+(yb-ya)^2')
xc =
xc-yb+yc
xc+yb-yc
-i*((xc+i*(yb-yc))*xc-xc^2+yb*yc-yc^2-yb*ya+ya*yc)/(yb-yc)
i*((xc-i*(yb-yc))*xc-xc^2+yb*yc-yc^2-yb*ya+ya*yc)/(yb-yc)
yc =
ya+xc-yc
-ya+yc+xc
xc+i*(yb-yc)
xc-i*(yb-yc)
请问各路大神,MatLab中我求二元二次函数,怎么会这样的结果,看不懂啊,
>> [xc,yc]=solve('2*(xa-xc)^2+2*(ya-yc)^2=(xb-xa)^2+(yb-ya)^2','2*(xb-xc)^2+2*(yb-yc)^2=(xb-xa)^2+(yb-ya)^2')
xc =
xc-yb+yc
xc+yb-yc
-i*((xc+i*(yb-yc))*xc-xc^2+yb*yc-yc^2-yb*ya+ya*yc)/(yb-yc)
i*((xc-i*(yb-yc))*xc-xc^2+yb*yc-yc^2-yb*ya+ya*yc)/(yb-yc)
yc =
ya+xc-yc
-ya+yc+xc
xc+i*(yb-yc)
xc-i*(yb-yc)
没错呀.结果是时域和代数.就是不知道a和b的值域.