原题:若方程|x²-5x|=a有且只有两相异实根,求a的取值范围解法:|x²-5x| = a ,则 x²-5x = ±a (a≥0);即 x²-5x-a = 0 或 x²-5x+a = 0 .当 a = 0 时,x²-5x+a = 0 与 x²-5x-a = 0 都变
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![原题:若方程|x²-5x|=a有且只有两相异实根,求a的取值范围解法:|x²-5x| = a ,则 x²-5x = ±a (a≥0);即 x²-5x-a = 0 或 x²-5x+a = 0 .当 a = 0 时,x²-5x+a = 0 与 x²-5x-a = 0 都变](/uploads/image/z/1768796-44-6.jpg?t=%E5%8E%9F%E9%A2%98%EF%BC%9A%E8%8B%A5%E6%96%B9%E7%A8%8B%EF%BD%9Cx%26sup2%3B-5x%EF%BD%9C%3Da%E6%9C%89%E4%B8%94%E5%8F%AA%E6%9C%89%E4%B8%A4%E7%9B%B8%E5%BC%82%E5%AE%9E%E6%A0%B9%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E8%A7%A3%E6%B3%95%EF%BC%9A%EF%BD%9Cx%26sup2%3B-5x%EF%BD%9C+%3D+a+%2C%E5%88%99+x%26sup2%3B-5x+%3D+%C2%B1a+%EF%BC%88a%E2%89%A50%EF%BC%89%EF%BC%9B%E5%8D%B3+x%26sup2%3B-5x-a+%3D+0+%E6%88%96+x%26sup2%3B-5x%2Ba+%3D+0+.%E5%BD%93+a+%3D+0+%E6%97%B6%2Cx%26sup2%3B-5x%2Ba+%3D+0+%E4%B8%8E+x%26sup2%3B-5x-a+%3D+0+%E9%83%BD%E5%8F%98)
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