在△ABC中,内角A B C所对的边分别是a b c,且a²+b²=c&#178

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∵ a²+b²=c²+ab
　　由余弦定理知,
　　cosC=(a^2 + b^2 - c^2)/(2ab)
　　= ab/(2ab) = 1/2
∵C是三角形内角,∴ C∈(0,π) , ∴C = π/3
∴A + B = 2π/3 ∴ B = 2π/3 - A
|M-2N|^2 = (sinA - 2cosB)^2 + (cosA + 2sinB)^2
= (sinA)^2 - 4sinAcosB + 4(cosB)^2 + (cosA)^2 + 4cosAsinB + 4（sinB)^2
=5 - 4(sinAcosB - cosAsinB)
=5 - 4sin(A - B)
=5 - 4sin(A - (2π/3 - A))
=5 - 4sin(2A - 2π/3)
∵C = π/3 , ∴A∈(0, 2π/3) ∴ (2A - 2π/3)∈( -2π/3,2π/3)
由正弦函数性质知,sin(2A - 2π/3)∈[-1, 1]
∴ |M-2N|^2 = 5 - 4sin(2A - 2π/3) ∈[1, 9]
又∵ |M-2N| > 0 ∴ |M-2N|∈[1, 3]
望采纳!