Bài I
\(1,\\ a,A=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{2}=\sqrt{3}+2-\sqrt{3}-1=1\\ b,B=\left(\tan20^0.\tan70^0\right).\left(\tan30^0.\tan60^0\right).\left(\tan40^0.\tan50^0\right)\\ B=\left(\tan20^0.\cot20^0\right).\left(\tan30^0.\cot30^0\right).\left(\tan40^0.\cot40^0\right)=1.1.1=1\\ 2,\\ ĐK:x\ge0\\ PT\Leftrightarrow\left|\sqrt{x}-2\right|=3\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=3\\2-\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=25\left(tm\right)\\\sqrt{x}=-1\left(ktm\right)\end{matrix}\right.\)
Bài II:
\(1,A=\dfrac{3\cdot3-4}{3+1}=\dfrac{5}{4}\\ 2,B=\dfrac{2x+\sqrt{x}-4-x+4+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ 3,P=AB=\dfrac{3\sqrt{x}-4}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{3\sqrt{x}-4}{\sqrt{x}-1}\\ P\ge2\Leftrightarrow\dfrac{3\sqrt{x}-4}{\sqrt{x}-1}-2\ge0\\ \Leftrightarrow\dfrac{3\sqrt{x}-4-2\sqrt{x}+2}{\sqrt{x}-1}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}\ge2\\\sqrt{x}< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\0\le x< 1\end{matrix}\right.\)
