Question

用配方法说明5x²-6x+2的值恒大于零.

Answer 1

解：5x²-6x+2=5（x²-6/5 x +2/5）
=5[(x-6/10)²+2/5-36/100]
=5(x-6/10)²+1/5
(x-6/10)²≥0
所以5(x-6/10)²+1/5＞0恒成立

Answer 2

y = 5x²-6x+2 = 5(x²-6/5 x) +2 = 5{ x²-6/5 x + (3/5)² -(3/5)² } +2 = 5(x-3/5)² - 5*(3/5)² +2
= 5(x-3/5)² + 1/5
∵5(x-3/5)² ≥0
∴ 5(x-3/5)² + 1/5 ≥1/5
∴5x²-6x+2的值恒大于零

Answer 3

5x²-6x+2＝5（x²－6/5x）+2
=5（x²－6/5x＋9/25-9/25）+2
=5[(x-3/5)²－9/25]+2
=5(x-3/5)²－9/5+2
=5(x-3/5)²+1/5
∵(x-3/5)²≥0
∴5(x-3/5)²+1/5>0恒成立。

Answer 4

5x²-6x+2=5(x²-6x/5+2/5)
=5(x²-6x/5+9/25-9/25+2/5)
=5(x²-6x/5+9/25)+1/5
=5(x-3/5)^2+1/5＞0

Answer 5

主要把它化成a(x+b)^2+c就行了，结果转化后C是一个大于0的数就成立。
5（X-3/5）^2+2-9/5=5（X-3/5）^2+1/5,因为5（X-3/5）^2>=0,
所以5（X-3/5）^2+1/5>=1/5>0
(这里a=5,b=-3/5,c=1/5)