\(B=-2\left(x^2-\dfrac{3}{2}x\right)+5=-2\left(x^2-2x.\dfrac{3}{4}+\dfrac{9}{16}\right)+5+\dfrac{9}{16}=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{25}{16}\le\dfrac{25}{16}\)
dấu = xảy ra khi x=3/4
vậy Bmax=....
tik mik nha
B=-2x2-3x+5
=-2(x2-\(\dfrac{3}{2}\)x)+5
=-2(x2-2.\(\dfrac{3}{4}\)x+\(\dfrac{9}{16}\))+\(\dfrac{71}{16}\)
=-2(x-\(\dfrac{3}{4}\))2+\(\dfrac{71}{16}\)≤\(\dfrac{71}{16}\)
Dấu "=" xảy ra⇔x-\(\dfrac{3}{4}\)=0⇔x=\(\dfrac{3}{4}\)
MaxB=\(\dfrac{71}{16}\)⇔×=\(\dfrac{3}{4}\)
Ta có: \(B=-2x^2-3x+5\)
\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)
\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)
