\(\begin{cases}\sqrt{xy}+\frac{1}{\sqrt{xy}}=\frac{5}{2}\\\sqrt{x}+\sqrt{y}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{9}{2}\end{cases}\)

<=>\(\begin{cases}xy+1=\frac{5\sqrt{xy}}{2}\\\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}+\sqrt{y}=\frac{9\sqrt{xy}}{2}\end{cases}\)

Đặt P=\(\sqrt{xy}\);S=\(\sqrt{x}+\sqrt{y}\)(S²\(\ge\)4P)

Ta có HPT: \(\begin{cases}P^2+1=\frac{5P}{2}\\S.P+P=\frac{9P}{2}\end{cases}\)

Tới đây dễ tự làm 

Khử mẫu đặt S P

là sao triều cậu giải thử đi 

Đặt \(a=\sqrt{2x-3}\) ; \(b=\sqrt{y-2}\) ; \(c=\sqrt{3z-1}\) (\(a,b,c>0\))

Ta có : \(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c=14\)

\(\Leftrightarrow\left(\sqrt{2x-3}+\frac{1}{\sqrt{2x-3}}-2\right)+\left(\sqrt{y-2}+\frac{4}{\sqrt{y-2}}-4\right)+\left(\sqrt{3z-1}+\frac{16}{\sqrt{3z-1}}-8\right)=0\)

\(\Leftrightarrow\left[\frac{\left(2x-3\right)-2\sqrt{2x-3}+1}{\sqrt{2x-3}}\right]+\left[\frac{\left(y-2\right)-4\sqrt{y-2}+4}{\sqrt{y-2}}\right]+\left[\frac{\left(3z-1\right)-8\sqrt{3z-1}+16}{\sqrt{3z-1}}\right]=0\)

\(\Leftrightarrow\frac{\left(\sqrt{2x-3}-1\right)^2}{\sqrt{2x-3}}+\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}+\frac{\left(\sqrt{3z-1}-4\right)^2}{\sqrt{3z-1}}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{2x-3}-1\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{3z-1}-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}}\)(TMĐK)

Vậy : \(\left(x;y;z\right)=\left(2;6;\frac{17}{3}\right)\)

Phần đặt ẩn a,b,c bạn bỏ đi nhé ^^

\(3=x+y+xy\le\sqrt{2\left(x^2+y^2\right)}+\dfrac{x^2+y^2}{2}\)

\(\Rightarrow\left(\sqrt{x^2+y^2}-\sqrt{2}\right)\left(\sqrt{x^2+y^2}+3\sqrt{2}\right)\ge0\)

\(\Rightarrow x^2+y^2\ge2\)

\(\Rightarrow-\left(x^2+y^2\right)\le-2\)

\(P=\sqrt{9-x^2}+\sqrt{9-y^2}+\dfrac{x+y}{4}\le\sqrt{2\left(9-x^2+9-y^2\right)}+\dfrac{\sqrt{2\left(x^2+y^2\right)}}{4}\)

\(P\le\sqrt{2\left(18-x^2-y^2\right)}+\dfrac{1}{4}.\sqrt{2\left(x^2+y^2\right)}\)

\(P\le\left(\sqrt{2}-1\right)\sqrt{18-x^2-y^2}+\sqrt[]{2}\sqrt{\dfrac{\left(18-x^2-y^2\right)}{2}}+\dfrac{1}{2}\sqrt{\dfrac{x^2+y^2}{2}}\)

\(P\le\left(\sqrt{2}-1\right).\sqrt{18-2}+\sqrt{\left(2+\dfrac{1}{4}\right)\left(\dfrac{18-x^2-y^2+x^2+y^2}{2}\right)}=\dfrac{1+8\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=y=1\)