\(A=x-x^2=-\left(x^2-x\right)=-\left(x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}-\frac{1}{4}\right)=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(A=x-x^2=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Vậy maxA = 1/4 <=> x = 1/2
A = x - x2
= -( x2 - x + 1/4 ) + 1/4
= -( x - 1/2 )2 + 1/4
\(-\left(x-\frac{1}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2
=> MaxA = 1/4 <=> x = 1/2
