\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right)\)
\(\div\left(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}+1\right)\left(1-\sqrt{xy}\right)+\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{xy}+1\right)\left(1-\sqrt{xy}\right)}{\left(\sqrt{xy}+1\right)\left(1-\sqrt{xy}\right)}\right]\)
\(\div\left[\dfrac{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)-\left(\sqrt{xy}+1\right)\left(\sqrt{x}+\sqrt{xy}\right)-\left(\sqrt{xy}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]\)
\(=\dfrac{2\left(\sqrt{x}+1\right)}{1-xy}\times\dfrac{xy-1}{-2\sqrt{xy}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{xy}}{xy}\)
Áp dụng BĐT AM - GM, ta có:
\(6=\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\ge2\times\sqrt{\dfrac{1}{\sqrt{xy}}}\)
\(\Leftrightarrow\sqrt{xy}\ge\dfrac{1}{9}\)
Ta có:
\(M=\dfrac{\sqrt{xy}}{xy}=\dfrac{1}{\sqrt{xy}}\le\dfrac{1}{\dfrac{1}{9}}=9\)
Max = 9 <=> x = y = \(\dfrac{1}{9}\)