Tìm Max a) P = 5 – 2x – x^2 Q= -3y^2 + 2y – 1

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

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