\(1,\\ a,ĐK:3-x\le0\Leftrightarrow x\ge3\\ b,B=\sqrt{\dfrac{x^2+1}{x}}\\ ĐK:x\ne0;x\ge0\Leftrightarrow x>0\\ 2,\\ a,M=2-\sqrt{6-2\sqrt{5}}+2\sqrt{5}\\ M=2-\sqrt{5}+1+2\sqrt{5}=3+\sqrt{5}\\ b,N=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)\\ N=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}-\sqrt{2}\right)^2\\ N=2\sqrt{10}-7\\ 3,\\ a,P=\dfrac{a\sqrt{a}-1}{\sqrt{a}-1}=a+\sqrt{a}+1\\ b,a=\dfrac{9}{4}\Leftrightarrow\sqrt{a}=\dfrac{3}{2}\\ \Leftrightarrow P=\dfrac{9}{4}+\dfrac{3}{2}+1=\dfrac{19}{4}\\ 4,\\ a,\Leftrightarrow\left|2x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\1-2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(b,ĐK:x\ge0\\ PT\Leftrightarrow3\sqrt{x}+11=8\sqrt{x}+1\\ \Leftrightarrow5\sqrt{x}=10\Leftrightarrow x=4\left(tm\right)\\ 5,\\ ĐK:x\le\dfrac{1}{3}\\ BPT\Leftrightarrow1-3x< 4\Leftrightarrow-3x< 3\Leftrightarrow x>-1\\ \Leftrightarrow-1< x\le\dfrac{1}{3}\)
