Question

求证:多项式(x2-4)(x2-10x+21)+100的值一定是非负数

Answer 1

(x2-4)(x2-10x+21)+100

=(x2-4)*[(x-5)2-4]+100

=x2*(x-5)2-4*(x-5)2-4*(x2-4)+16+100 化简得

=x2*(x-5)2-8*x*(x-5)+16 这步是关键

=[x*(x-5)-4]2

一个平方数必大于等于0

*表示乘号;