已知二次函数f(x)=ax²+bx+c,若x1<x2,且f(x1)≠f(x2)求证关于x的方程f(x)=1/2[f(x1)+f(x2)]在区间(x1,x2)内必有一根
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![已知二次函数f(x)=ax²+bx+c,若x1<x2,且f(x1)≠f(x2)求证关于x的方程f(x)=1/2[f(x1)+f(x2)]在区间(x1,x2)内必有一根](/uploads/image/z/468763-43-3.jpg?t=%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dax%26%23178%3B%2Bbx%2Bc%2C%E8%8B%A5x1%EF%BC%9Cx2%2C%E4%B8%94f%EF%BC%88x1%EF%BC%89%E2%89%A0f%EF%BC%88x2%EF%BC%89%E6%B1%82%E8%AF%81%E5%85%B3%E4%BA%8Ex%E7%9A%84%E6%96%B9%E7%A8%8Bf%EF%BC%88x%EF%BC%89%3D1%2F2%EF%BC%BBf%EF%BC%88x1%EF%BC%89%2Bf%EF%BC%88x2%EF%BC%89%EF%BC%BD%E5%9C%A8%E5%8C%BA%E9%97%B4%EF%BC%88x1%2Cx2%EF%BC%89%E5%86%85%E5%BF%85%E6%9C%89%E4%B8%80%E6%A0%B9)
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