\(x-x^2-1=-\left(x^2-x+1\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\Leftrightarrow-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\le\frac{3}{4}< 0\)
Vậy x-x^2 -1 < 0
Ta có : \(x^2-x+1\)
\(=-x^2+x-1\)
\(=-\left(x^2-x+1\right)\)
\(=-\left[x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]+\frac{3}{4}\)
\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{-3}{4}\)
Ta có : \(-\left(x-\frac{1}{2}\right)^2\le0\)
\(\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{-3}{4}\le-\frac{3}{4}\)
Dấu " = " xảy ra khi và chỉ khi \(x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Max\) của biểu thức là \(-\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
