a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=1\\x\notin\left\{3;2\right\}\end{matrix}\right.\)
b: \(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\)
\(=\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\dfrac{x-1-2}{\sqrt{x-1}-\sqrt{2}}\)
\(=\sqrt{x}+\sqrt{x-1}-\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\)
\(=\sqrt{x}+\sqrt{x-1}-\left(\sqrt{x-1}+\sqrt{2}\right)=\sqrt{x}-\sqrt{2}\)
\(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\)
\(=\dfrac{-2}{\sqrt{x}-\sqrt{2}}+\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{x}-\sqrt{2}\right)}\)
\(=\dfrac{-2\sqrt{x}+\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{x}-\sqrt{2}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{x}-\sqrt{2}\right)}=\dfrac{-1}{\sqrt{x}}\)
\(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{-1}{\sqrt{x}}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Thay \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) vào P, ta được:
\(P=\dfrac{-\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{-\sqrt{2}-1+\sqrt{2}}{\sqrt{2}+1}=\dfrac{-1}{\sqrt{2}+1}\)
\(=\dfrac{-\left(\sqrt{2}-1\right)}{2-1}=-\sqrt{2}+1\)