\(B=-\left(x^2-3x-1\right)\)
\(=-\left(x^2-3x+\dfrac{9}{4}-\dfrac{13}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{13}{4}\le\dfrac{13}{4}\)
Dấu '=' xảy ra khi x=3/2
\(B=1-x^2+3x\)
\(B=-\left(x^2-3x\right)+1\)
\(B=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{9}{4}+1\)
\(B=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{13}{4}\le\dfrac{13}{4}\)
Vậy Max B = \(\dfrac{13}{4}\) đạt tại x = \(\dfrac{3}{2}\)
`B=-(x^2-3x)+1`
`=-(x^2-3x+9/4)+9/4+1`
`=-(x-3/2)^2+13/4<=13/4`
Vậy `max B=13/4` khi `x=3/2`
B = 1 - x2 + 3X
B = - (x2 - 2.\(\dfrac{3}{2}\)x + \(\dfrac{9}{4}\)) + \(\dfrac{13}{4}\)
B = - (x - \(\dfrac{3}{2}\))2 + \(\dfrac{13}{4}\)
- ( x - 3/2)2 \(\le\) 0
B = -(x -3/2) + 13/4 \(\le\)13/4
B max = 13/4 dấu bằng xảy ra khi x - 3/2 = 0 \(\Rightarrow\) x = 3/2
