Trả lời chủ quan bị cộng sai:
\(A=-x^2+x+1=\left(1+\frac{1}{4}\right)-\left(x^2-2.\frac{1}{2}+\frac{1}{4}\right)\) --> thêm 1/4 bên ngoài (trừ 1/4 trong ngoặc)
\(A=\frac{5}{4}-\left(x-\frac{1}{2}\right)^2\)
ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\frac{1}{2}\right)^2\le0\Rightarrow\frac{5}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{5}{4}\)
Vậy GTLN A=5/4 khi x=1/2.
A=\(-x^2+x+1=-x^2+2\cdot\frac{1}{2}x-\frac{1}{4}+\frac{1}{4}+1\\ =-\left(x^2-2\cdot\frac{1}{2}x+\frac{1}{4}\right)+\frac{5}{4}=-\left(x+\frac{1}{2}\right)^2\le\frac{5}{4}\)
Dấu "=" xảy ra khi \(-\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Vậy GTLN của A là 5/4 tại x=1/2
\(A=-x^2+x+1=-\left(x^2-2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2\right)+\frac{1}{4}=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)
GTLN A=1/4 khi x=1/2
