\(x-x^2+\dfrac{1}{4}\)
\(=-\left(x^2-x-\dfrac{1}{4}\right)\)
\(=-\left[\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{1}{4}-\dfrac{1}{4}\right]\)
\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)
= \(-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)
Ta có :
\(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\)
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\) ≤ \(\dfrac{1}{2}< 0\)
Đúng 0
Bình luận (0)
