Từ x-y=2=>x=y+2

a)Thay x=y+2 vào P ta có:

\(P=xy+4=\left(y+2\right)y+4=y^2+2y+4=\left(y^2+2y+1\right)+3=\left(y^2+2.y.1+1^2\right)+3\)

\(=\left(y+1\right)^2+3\ge3\) với mọi y

Dấu "=" xảy ra <=> \(\left(y+1\right)^2=0\) <=> \(y=-1\) <=> \(x=1\)

Vậy...........

b)Thay x=y+2 vào Q ta có:

\(Q=x^2+y^2-xy=\left(y+2\right)^2+y^2-\left(y+2\right).y=y^2+4y+4+y^2-y^2-2y\)

\(=y^2-2y+4=\left(y^2-2y+1\right)+3=\left(y^2-2.y.1+1^2\right)+3=\left(y-1\right)^2+3\ge3\) với mọi y

Dấu "=" xảy ra <=> y=1 <=> x=2

Vậy.................

\(Q=\frac{x^2+1,2xy+y^2}{x-y}=\frac{x^2-2xy+y^2+3,2xy}{x-y}\)

\(=\frac{\left(x-y\right)^2+48}{x-y}=\frac{\left(x-y\right)^2}{x-y}+\frac{48}{x-y}\)

\(=x-y+\frac{48}{x-y}\ge2\sqrt{48}=8\sqrt{3}\)

1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)

 \(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)

max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t

\(A=x^2+y^2=\frac{\left(1^2+1^2\right)\left(x^2+y^2\right)}{2}\ge\frac{\left(1.x+1.y\right)^2}{2}=\frac{1}{2}\)A min = 1 khi x =y = 1/2

\(\sqrt{A}=\sqrt{x^2+y^2}\le\sqrt{x^2}+\sqrt{y^2}=x+y=1\)( \(\sqrt{a+b}\le\sqrt{a}+\sqrt{b}\))

=> A\(\le1\) => Max A = 1 khi x =0;y =1 hoặc x =1 ; y =0

\(\Rightarrow Q=\frac{\left(x-y\right)^2+14xy}{x-y}\) mà xy=5

\(\Rightarrow Q=\frac{\left(x-y\right)^2+70}{x-y}\)mà x>y nên x-y lớn hơn 0.

Áp dụng Cô si ta có : \(Q\ge\frac{2.\sqrt{\left(x-y\right)^2.70}}{x-y}=\frac{2.\sqrt{70}.\left(x-y\right)}{x-y}=2.\sqrt{70}\)

Dấu = xảy ra thì thế vào là được.