求证:多项式(x2-4﹚﹙x2-10x+21﹚+100的值一定是非负数
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/22 17:28:24
![求证:多项式(x2-4﹚﹙x2-10x+21﹚+100的值一定是非负数](/uploads/image/z/6853731-51-1.jpg?t=%E6%B1%82%E8%AF%81%3A%E5%A4%9A%E9%A1%B9%E5%BC%8F%28x2-4%EF%B9%9A%EF%B9%99x2-10x%2B21%EF%B9%9A%2B100%E7%9A%84%E5%80%BC%E4%B8%80%E5%AE%9A%E6%98%AF%E9%9D%9E%E8%B4%9F%E6%95%B0)
求证:多项式(x2-4﹚﹙x2-10x+21﹚+100的值一定是非负数
求证:多项式(x2-4﹚﹙x2-10x+21﹚+100的值一定是非负数
求证:多项式(x2-4﹚﹙x2-10x+21﹚+100的值一定是非负数
证明:
(x²-4﹚﹙x²-10x+21﹚+100
=(x+2)(x-2)(x-3)(x-7)+100
=(x²-5x-14)(x²-5x+6)+100
=(x²-5x)²+6(x²-5x)-14(x²-5x)-84+100
=(x²-5x)²-8(x²-5x)+16
=(x²-5x-4)²
因为(x²-5x-4)²是一非负数,所以多项式(x2-4﹚﹙x2-10x+21﹚+100的值一定是非负数.
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