1: \(A=\left(\dfrac{\sqrt{x}-1}{x-4}-\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}\right):\dfrac{x\sqrt{x}}{\left(4-x\right)^2}\)
\(=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{\left(x-4\right)^2}{x\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\left(\sqrt{x}-2\right)^2\cdot\left(\sqrt{x}+2\right)^2}{x\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{1}\cdot\dfrac{\sqrt{x}-2}{x\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{x\sqrt{x}}\cdot\left(\sqrt{x}-2\right)=\dfrac{2\left(\sqrt{x}-2\right)}{x}\)
2: Thay \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\) vào A, ta được:
\(A=\dfrac{2\left[\sqrt{\left(\sqrt{3}-1\right)^2}-2\right]}{4-2\sqrt{3}}\)
\(=\dfrac{2\left(\sqrt{3}-1-2\right)}{\left(\sqrt{3}-1\right)^2}=\dfrac{2\left(\sqrt{3}-3\right)}{\left(\sqrt{3}-1\right)^2}\)
\(=\dfrac{-2\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)^2}=\dfrac{-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{-2\sqrt{3}\left(\sqrt{3}+1\right)}{3-1}=-\sqrt{3}\left(\sqrt{3}+1\right)\)
3: Để A>=1/4 thì \(\dfrac{2\left(\sqrt{x}-2\right)}{x}-\dfrac{1}{4}>=0\)
=>\(\dfrac{8\left(\sqrt{x}-2\right)-x}{4x}>=0\)
=>\(-x+8\sqrt{x}-16>=0\)
=>\(\left(\sqrt{x}-4\right)^2< =0\)
=>\(\sqrt{x}-4=0\)
=>x=16
