mathematica怎么求解方程组Reduce[{s1^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]])^2 + (r Sin[\[Alpha]] \Cos[\[Beta]] + d/2)^2 + (h - r Cos[\[Alpha]])^2,s2^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]] - l)^2 + (r Sin[\[Alpha]] Cos[\[Beta]] - d/2)^2 + (h -
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![mathematica怎么求解方程组Reduce[{s1^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]])^2 + (r Sin[\[Alpha]] \Cos[\[Beta]] + d/2)^2 + (h - r Cos[\[Alpha]])^2,s2^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]] - l)^2 + (r Sin[\[Alpha]] Cos[\[Beta]] - d/2)^2 + (h -](/uploads/image/z/6363264-48-4.jpg?t=mathematica%E6%80%8E%E4%B9%88%E6%B1%82%E8%A7%A3%E6%96%B9%E7%A8%8B%E7%BB%84Reduce%5B%7Bs1%5E2+%3D%3D+%28x+%2B+r+Sin%5B%5C%5BAlpha%5D%5D+Sin%5B%5C%5BBeta%5D%5D%29%5E2+%2B+%28r+Sin%5B%5C%5BAlpha%5D%5D+%5CCos%5B%5C%5BBeta%5D%5D+%2B+d%2F2%29%5E2+%2B+%28h+-+r+Cos%5B%5C%5BAlpha%5D%5D%29%5E2%2Cs2%5E2+%3D%3D+%28x+%2B+r+Sin%5B%5C%5BAlpha%5D%5D+Sin%5B%5C%5BBeta%5D%5D+-+l%29%5E2+%2B+%28r+Sin%5B%5C%5BAlpha%5D%5D+Cos%5B%5C%5BBeta%5D%5D+-+d%2F2%29%5E2+%2B+%28h+-)
mathematica怎么求解方程组Reduce[{s1^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]])^2 + (r Sin[\[Alpha]] \Cos[\[Beta]] + d/2)^2 + (h - r Cos[\[Alpha]])^2,s2^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]] - l)^2 + (r Sin[\[Alpha]] Cos[\[Beta]] - d/2)^2 + (h -
mathematica怎么求解方程组
Reduce[{s1^2 == (x +
r Sin[\[Alpha]] Sin[\[Beta]])^2 + (r Sin[\[Alpha]] \
Cos[\[Beta]] + d/2)^2 + (h - r Cos[\[Alpha]])^2,
s2^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]] -
l)^2 + (r Sin[\[Alpha]] Cos[\[Beta]] - d/2)^2 + (h -
r Cos[\[Alpha]])^2,0 < \[Alpha] < Pi/2,
0 < \[Beta] < Pi/2},{\[Alpha],\[Beta]}]
用 mathematica 始终在运行啊
mathematica怎么求解方程组Reduce[{s1^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]])^2 + (r Sin[\[Alpha]] \Cos[\[Beta]] + d/2)^2 + (h - r Cos[\[Alpha]])^2,s2^2 == (x + r Sin[\[Alpha]] Sin[\[Beta]] - l)^2 + (r Sin[\[Alpha]] Cos[\[Beta]] - d/2)^2 + (h -
解析解太复杂,无法求出啊.只能将未知数的实际值代入方程,求得数值解.