Question

tìm GTLN của x - x² +1.

cần gấp ạ.

Answer 1

\(x-x^2+1=-x^2+x+1=-\left(x^2-x-1\right)=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)

Có: \(\left(x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\le-\frac{3}{4}\)

Dấu = xảy ra khi:

\(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

Vậy:...............