1: ĐKXĐ: x>1; x<>10
\(P=\left(\dfrac{\sqrt{x-1}}{3+\sqrt{x-1}}+\dfrac{x+8}{10-x}\right):\left(\dfrac{3\sqrt{x-1}+1}{x-3\sqrt{x-1}-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\left(\dfrac{\sqrt{x-1}}{\sqrt{x-1}+3}-\dfrac{x+8}{\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+3\right)}\right):\left(\dfrac{3\sqrt{x-1}+1}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\dfrac{\sqrt{x-1}\left(\sqrt{x-1}-3\right)-x-8}{\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+3\right)}:\dfrac{3\sqrt{x-1}+1-\sqrt{x-1}+3}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)
\(=\dfrac{x-1-3\sqrt{x-1}-x-8}{\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+3\right)}\cdot\dfrac{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}{2\sqrt{x-1}+4}\)
\(=\dfrac{-3\left(\sqrt{x-1}+3\right)}{\sqrt{x-1}+3}\cdot\dfrac{\sqrt{x-1}}{2\left(\sqrt{x-1}+2\right)}\)
\(=\dfrac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)
2:
\(x=\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)
\(=\sqrt[4]{\left(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\right)^2}-\sqrt[4]{\left(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\right)^2}\)
\(=\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}-\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{2-1}}-\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1-\sqrt{2}+1=2\)
Khi x=2 thì \(P=\dfrac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=\dfrac{-3}{2+4}=-\dfrac{1}{2}\)
