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 $x=\dfrac{1}{2}$

 

Đáp án:

a,M = 4x - x² + 3

       = - (x² - 4x - 3)

       = - (x² - 2.2x + 2² - 7)

       = - (x - 2)² + 7 ≤ 7

Mmax = 7 ⇔ (x - 2)² = 0

                 ⇔   x - 2 = 0

                 ⇔   x = 2

Vậy Mmax = 7 khi x = 2

b, N = x - x²

        = - (x² - x)

        = - (x² - 2.$\frac{1}{2}$x + $\frac{1}{4}$ - $\frac{1}{4}$ )

        = - ( x - $\frac{1}{2}$)² + $\frac{1}{4}$ ≤ $\frac{1}{4}$

N max = $\frac{1}{4}$ ⇔ (x - $\frac{1}{2}$)² = 0

                                    ⇔ x - $\frac{1}{2}$ = 0

                                    ⇔ x = $\frac{1}{2}$

Vậy N max = $\frac{1}{4}$ khi x = $\frac{1}{2}$

c, P = 2x - 2x² - 5

      = - 2(x² - x + $\frac{5}{2}$ )

      = - 2(x² - x + $\frac{1}{4}$ + $\frac{9}{4}$)

      = - 2(x - $\frac{1}{2}$)² - $\frac{9}{2}$ ≤ -$\frac{9}{2}$

   Pmax =- $\frac{9}{2}$ ⇔ (x - $\frac{1}{2}$)² = 0

                                      ⇔ x = $\frac{1}{2}$

Vậy Pmax =- $\frac{9}{2}$ khi x =$\frac{1}{2}$