L = 1.105;
a = 0.18394;
R = 0.5;
Solve[x^2/(L/2)^2 + x^2*(Tan[18°]])^2/(a/2)^2 == 1]
ym = xm*Tan[18°]]
m = 2*Sqrt[ym^2 + xm^2]
hmax = R - Sqrt[R^2 - ((L*Sin[18°]])/2)^2]
Solve[R1p - Sqrt[R1p^2 - (m/2)^2] == hmax]
Solve[x^2/(L/2)^2 + x^2*(Tan[18°]])^2/(a/2)^2 == 1]
这句改成
xm = x /.
Solve[x^2/(L/2)^2 + x^2*(Tan[18 \[Degree]])^2/(a/2)^2 == 1][[1]]
试试,不困脊悔知道你问的汪正是不是这个意思?题目表野清述的不清楚.
这个胡锋敏基行我可以做裤枝的,http://hi.baidu.com/ggggwhw/blog/item/1ed2d35d5130b95efaf2c07b.html