\(•B=4x-9x^2=-\left(9x^2-4x\right)\\ =-\left(9x^2-3x.2.\dfrac{2}{3}+\dfrac{4}{9}\right)+\dfrac{4}{9}\\ =-\left(3x-\dfrac{2}{3}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\\dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{2}{9}\\ vậy\: MAX_B=\dfrac{4}{9}\: tại\: x=\dfrac{2}{9}\\ •C=5-2x-4x^2=-\left(4x^2+2x-5\right)\\ =-\left(4x^2+2.2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{21}{4}\\ =-\left(2x+\dfrac{1}{2}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=-\dfrac{1}{4}\\ vậy\: MAX_C=\dfrac{21}{4}\: tại\: x=\dfrac{-1}{4}\\ •D=7+3x-x^2=-\left(x^2-3x-7\right)\\ =-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{37}{4}\\ =-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{3}{2}\\ vậy\: MAX_D=\dfrac{37}{4}\: tại\: x=\dfrac{3}{2}\)\(•E=1+x-x^2=-\left(x^2-x-1\right)\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\\ dấu\:"="\:xảy\:ra\:khi\:x=\dfrac{1}{2}\\ vậy\:MAX_E=\dfrac{5}{4}\:tại\:x=\dfrac{1}{2}\\ •F=-5x-6x^2\\ -\dfrac{F}{6}=x^2+\dfrac{5}{6}x=x^2+2.x.\dfrac{5}{12}+\dfrac{25}{144}-\dfrac{25}{144}\\ -\dfrac{F}{6}=\left(x+\dfrac{5}{12}\right)^2-\dfrac{25}{144}\\ F=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{-5}{12}\\ vậy\: MAX_F=\dfrac{25}{24}\: tại\: x=\dfrac{-5}{12}\)
\(B=4x-9x^2=-9\left(x^2-\dfrac{4}{9}x+\dfrac{4}{81}\right)+\dfrac{4}{9}\)
\(=-9\left(x-\dfrac{2}{9}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\forall x\)
vậy Max B = \(\dfrac{4}{9}\) khi \(x-\dfrac{2}{9}=0\Rightarrow x=\dfrac{2}{9}\)
\(C=5-2x-4x^2=-4\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)+\dfrac{21}{4}\)\(=-4\left(x+\dfrac{1}{4}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\)
Vậy Max C = \(\dfrac{21}{4}\) khi \(x+\dfrac{1}{4}=0\Rightarrow x=-\dfrac{1}{4}\)
\(D=7+3x-x^2\)
\(=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{37}{4}\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\forall x\)
Vậy Max D = \(\dfrac{37}{4}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(E=1+x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{5}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\forall x\)
Vậy Max E = \(\dfrac{5}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(F=-5x-6x^2=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{25}{24}\)\(=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\forall x\)
Vậy Max F = \(\dfrac{25}{24}\) khi \(x+\dfrac{5}{12}=0\Leftrightarrow x=-\dfrac{5}{12}\)
