Question

B =x-x^2 tìm max

Answer 1

Ta có:\(B=-\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Dấu "=: xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

Answer 2

-( x^2 - 2x1/2 + 1/4) + 1/4

-(x - 1/2)^2 + 1/4 <=1/4

MaxB= 1/4 khi x=1/2

Answer 3

B=x-x² 

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

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

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

Ta có ;

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