Bài 1:
\(1,B=\dfrac{1}{4-3}=1\\ 2,M=A:B=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\left(\sqrt{x}-3\right)\\ M=\dfrac{x+7}{\sqrt{x}+3}\\ 3,M=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6\\ M\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{16}{\sqrt{x}+3}}-6=2\\ M_{min}=6\Leftrightarrow\sqrt{x}+3=4\Leftrightarrow x=1\left(tm\right)\)
Bài 2:
\(1,A=\dfrac{1-3}{1+3}=-\dfrac{1}{2}\\ 2,B=\dfrac{6-\sqrt{x}+2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\\ B=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\\ 3,B-A\le\dfrac{3}{5}\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{1+\sqrt{x}}-\dfrac{3}{5}\le0\\ \Leftrightarrow\dfrac{5-5+5\sqrt{x}-3\sqrt{x}-3}{5\left(\sqrt{x}+1\right)}\le0\\ \Leftrightarrow2\sqrt{x}-3\le0\left(\sqrt{x}+1>0\right)\\ \Leftrightarrow\sqrt{x}< \dfrac{3}{2}\\ \Leftrightarrow0\le x\le\dfrac{9}{4}\)
