Tìm giá trị của biến để biểu thức đạt giá trị lớn nhất, giá trị nhỏ nhất

a, Q = $\left [ \frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}(\sqrt{y}+1)}{1-\sqrt{xy}}+1 \right ]:\left [ 1-\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{x}(\sqrt{y}+1)}{\sqrt{xy}-1} \right ]$ với $x\geq 0, y\geq 0$ và $xy\neq 1$

<=> Q = $\frac{(\sqrt{x}+1)( 1-\sqrt{xy})+\sqrt{x}(\sqrt{y}+1)( \sqrt{xy}+1)+ (\sqrt{xy}+1)(1-\sqrt{xy})}{ (\sqrt{xy}+1)(1-\sqrt{xy})}$

: $\frac{ (\sqrt{xy}+1)(\sqrt{xy}-1)-(\sqrt{x}+1)(\sqrt{xy}-1)- \sqrt{x}(\sqrt{y}+1)(\sqrt{xy}+1)}{ (\sqrt{xy}+1)(\sqrt{xy}-1)}$

<=> Q = $\frac{\sqrt{x}-\sqrt{x^{2}y}+1-\sqrt{xy}+xy+\sqrt{x}+\sqrt{x^{2}y}+\sqrt{xy}+1-xy}{1-xy}:\frac{xy-1-(\sqrt{x^{2}y}-\sqrt{x}-\sqrt{xy}-1)-(xy+\sqrt{x}+\sqrt{x^{2}y}+\sqrt{xy})}{xy-1}$

<=> Q = $\frac{2\sqrt{x}+2}{1-xy}:\frac{-2\sqrt{x^{2}y}-2\sqrt{xy}}{xy-1}$

<=> Q = $\frac{2.(\sqrt{x}+1)}{1-xy}.\frac{1-xy}{2\sqrt{xy}.(\sqrt{x}+1)}$ = $\frac{1}{\sqrt{xy}}$

Vậy Q = $\frac{1}{\sqrt{xy}}$

b, $x=2(3-\sqrt{5}),y=2(3+\sqrt{5})$

=> xy =  $2.(3-\sqrt{5}).2.(3+\sqrt{5})=4.(9-5)=16$ => $\sqrt{xy}=4$

Thay vào Q ta có: Q= $\frac{1}{4}$

c, $\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}=5$ <=> $\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{5}{2}$

<=> $\left ( \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}} \right )^{2}=\frac{25}{4}$ <=> $\frac{1}{x}+\frac{1}{y}=\frac{25}{4}-\frac{2}{\sqrt{xy}}$

Ta có: $\frac{1}{x}+\frac{1}{y}\geq 2.\frac{1}{\sqrt{x}}.\frac{1}{\sqrt{y}}$

<=> $\frac{1}{x}+\frac{1}{y}\geq 2.\frac{1}{\sqrt{xy}}$ <=> $\frac{25}{4}-\frac{2}{\sqrt{xy}}\geq \frac{2}{\sqrt{xy}}$

<=> $\frac{4}{\sqrt{xy}}\leq \frac{25}{4}$ <=> $\frac{1}{\sqrt{xy}}\leq \frac{25}{16}$ <=> $Q\leq \frac{25}{16}$

Vậy Q nhận GTLN bằng $\frac{25}{16}$ <=> $\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{y}}$ <=> x = y = $\frac{16}{25}$