Question

Answer 1

\(a,A=\dfrac{\sqrt{10}+3}{1}-\sqrt{\left(\sqrt{10}-3\right)^2}=\sqrt{10}+3-\sqrt{10}+3=6\\ B=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,A>9B\Leftrightarrow A-9B>0\\ \Leftrightarrow6-\dfrac{0\sqrt{x}-9}{\sqrt{x}}>0\Leftrightarrow\dfrac{6\sqrt{x}-9\sqrt{x}+9}{\sqrt{x}}>0\\ \Leftrightarrow9-3\sqrt{x}>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0< x< 9;x\ne1\)