`a)`

`P = 5 - 2x - x^2`

` = - (x^2 + 2x - 5)`

` = - (x^2 + 2x + 1)  + 6`

` = -( x^2 + 2 . x . 1 + 1^2) + 6`

` = - (x+1)^2 + 6`

`\forall x ` ta có :

`(x+1)^2 \ge 0`

`=> -(x+1)^2 \le 0`

`=> -(x+1)^2 + 6 \le 6`

`=> P \le 6`

Dấu `=` xảy ra `<=> x + 1 = 0`

`<=> x = -1`

Vậy `Max_P = 6 <=> x = -1`

`b)`

`Q = -3y^2 + 2y  -1`

`= -3 . (y^2 - 2/3y )  -1`

` = -3 . (y^2 - 2/3y + 1/9)  - 2/3`

` = -3 . [y^2 - 2 . y . 1/3 + (1/3)^2 ]  -2/3`

` = -3 . (y-1/3)^2 - 2/3`

`\forall x` ta có :

`(y-1/3)^2 \ge 0`

`=> -3 . (y-1/3)^2 \le 0`

`=> -3 . (y-1/3)^2 - 2/3 \le -2/3`

`=> Q \le -2/3`

Dấu `=` xảy ra `<=> y - 1/3 =0`

`<=> y=1/3`

Vậy `Max_Q = -2/3 <=> y = 1/3`

$a,P=5-2x-x^2 \\=-(x^2+2x-5) \\=-(x^2+2x+1-6) \\=-(x+1)^2+6 \\Vì\ -(x+1)^2≤0∀x \\⇒P=-(x+1)^2+6≤6∀x$

Dấu = xảy ra khi và chỉ khi 

$-(x+1)^2=0 \\⇔-(x+1)=0 \\⇔x+1=0 \\⇔x=-1 \\Vậy\ Max\ P=6⇔x=-1 \\b,Q=-3y^2+2y-1 \\=-(3y^2-2y+1) \\=-\bigg[(\sqrt{3}y)^2-2.\sqrt{3}y.\dfrac{1}{\sqrt{3}}+\bigg(\dfrac{1}{\sqrt{3}}\bigg)^2+\dfrac{2}{3}\bigg] \\=-\bigg[\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2+\dfrac{2}{3}\bigg] \\=-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-\dfrac{2}{3} \\Vì\ -\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-≤0∀x \\⇒Q=-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-\dfrac{2}{3}≤-\dfrac{2}{3}∀x$

Dấu = xảy ra khi và chỉ khi 

$-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2=0 \\⇔\sqrt{3}y-\dfrac{1}{\sqrt{3}}=0 \\⇔\sqrt{3}y=\dfrac{1}{\sqrt{3}} \\⇔y=\dfrac{1}{3}$

Vậy $Max\ Q=\dfrac{-2}{3}⇔y=\dfrac{1}{3}$