Đáp án:
$a) M = 4x -x^2+3$
$= -(x^2 -4x-3)$
$ = -(x^2 -2.x.2 +4-7)$
$ = -(x-2)^2 +7$
Vì $-(x-2)^2 ≤ 0$
Nên $-(x-2)^2 +7 ≤ 7$
Dấu''='' xảy ra khi $x-2 =0 ⇔x=2$
Vậy Max M = 7 tại x= 2
$b) N = x-x^2$
$ = -(x^2 -x+0)$
$ = -(x^2 -2.x.\dfrac{1}{2} +\dfrac{1}{4} -\dfrac{1}{4})$
$ = -(x-\dfrac{1}{2})^2 +\dfrac{1}{4}$
Vì $-(x-\dfrac{1}{2})^2 ≤ 0$
Nên $-(x-\dfrac{1}{2})^2 +\dfrac{1}{4} ≤ \dfrac{1}{4}$
Dấu ''='' xảy ra khi $x-\dfrac{1}{2} =0 ⇔x=\dfrac{1}{2}$
Vậy Max N = $\dfrac{1}{4}$ tại x=$\dfrac{1}{2}$
$c) P = 2x-2x^2 -5$
$ = -(2x^2 -2x+5)$
$ = -(√2x - 2 .√2x . \dfrac{\sqrt[]{2}}{2} + \dfrac{1}{2} + \dfrac{9}{2})$
$ = -(√2x-\dfrac{\sqrt[]{2}}{2})^2 -\dfrac{9}{2}$
Vì $-(√2x-\dfrac{\sqrt[]{2}}{2})^2 ≤ 0$
Nên $-(√2x-\dfrac{\sqrt[]{2}}{2})^2 -\dfrac{9}{2} ≤ -\dfrac{9}{2}$
Dấu ''='' xảy ra khi $ √2x-\dfrac{\sqrt[]{2}}{2} = 0⇔ x=\dfrac{1}{2}$
Vậy Max P = $-\dfrac{9}{2}$ tại $x=\dfrac{1}{2}$
