\(B=x-x^2\)

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

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

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

Ta có :

\(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Dấu = xảy ra \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(Max_B=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

Tìm giá trị lớn nhất biểu thức:

\(A=4x-x^2+3\)

\(A=4x-x^2+3\)

\(=-\left(x^2-4x+4\right)+7\)

\(=-\left(x-2\right)^2+7\)

Ta có :

\(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2+7\le7\)

Dấu = xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(Max_A=7\Leftrightarrow x=2\)