Question

[Nguyễn Đức Duy]

Chứng minh rằng:

\(x-x^2+\dfrac{1}{4}\le0\)

Answer 1

[lê thị hương giang]

\(x-x^2+\dfrac{1}{4}\)

\(=-\left(x^2-x-\dfrac{1}{4}\right)\)

\(=-\left[\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{1}{4}-\dfrac{1}{4}\right]\)

\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)

= \(-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)

Ta có :

\(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\)

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\) ≤ \(\dfrac{1}{2}< 0\)