设实数m,n满足4m^2+n^2=8,求√(m^2+n^2-4n+4)+√(m^2+n^2-4m-4n+8)的最小值.
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![设实数m,n满足4m^2+n^2=8,求√(m^2+n^2-4n+4)+√(m^2+n^2-4m-4n+8)的最小值.](/uploads/image/z/5548273-25-3.jpg?t=%E8%AE%BE%E5%AE%9E%E6%95%B0m%2Cn%E6%BB%A1%E8%B6%B34m%5E2%2Bn%5E2%3D8%2C%E6%B1%82%E2%88%9A%EF%BC%88m%5E2%2Bn%5E2-4n%2B4%29%2B%E2%88%9A%28m%5E2%2Bn%5E2-4m-4n%2B8%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC.)
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设实数m,n满足4m^2+n^2=8,求√(m^2+n^2-4n+4)+√(m^2+n^2-4m-4n+8)的最小值.
设实数m,n满足4m^2+n^2=8,求√(m^2+n^2-4n+4)+√(m^2+n^2-4m-4n+8)的最小值.
设实数m,n满足4m^2+n^2=8,求√(m^2+n^2-4n+4)+√(m^2+n^2-4m-4n+8)的最小值.
根号(m^2+n^2-4n+4)+根号(m^2+n^2-4m-4n+8)
=根号[m^2+(n-2)^2]+根号[(m-2)^2+(n-2)^2]
由以上形式,可将题目可看成是求点(m,n)到(0,2)和(2,2)两点距离之和的最小值
由坐标图上可看出,当n=2且0
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