1: ĐKXĐ: x>=0; x<>4
\(A=\dfrac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-x\sqrt{x}-2x}{x\sqrt{x}-8}\cdot\dfrac{x+2\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{4\sqrt{x}}{x-4}\)
2: Để A=1/2 thì \(\dfrac{4\sqrt{x}}{x-4}=\dfrac{1}{2}\)
\(\Leftrightarrow x-4=8\sqrt{x}\)
\(\Leftrightarrow x-8\sqrt{x}-4=0\)
\(\Leftrightarrow x=\left(4+2\sqrt{5}\right)^2\)
\(ĐK:\left\{{}\begin{matrix}x>=0\\\sqrt{x}-2\ne0\\x\sqrt{x}-8\ne0\\\sqrt{x}+2\ne0\end{matrix}\right.< =>x>=0;x\ne4\)
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x\left(\sqrt{x}+2\right)}{x\sqrt{x}-8}\right).\left(\dfrac{x}{\sqrt{x}+2}+2\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\right).\dfrac{x+2\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}.\dfrac{x+2\sqrt{x}+4}{\sqrt{x}+2}\\ =\dfrac{x\sqrt{x}+2x+4\sqrt{x}-x\sqrt{x}-2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{4\sqrt{x}}{x-4}\)
\(A=\dfrac{1}{2}< =>\dfrac{4\sqrt{x}}{x-4}=\dfrac{1}{2}\\ < =>x-4=8\sqrt{x}< =>x-8\sqrt{x}-4=0\\ < =>\left(\sqrt{x}-4\right)^2-20=0\\ =>\left[{}\begin{matrix}\sqrt{x}-4=\sqrt{20}\\\sqrt{x}-4=-\sqrt{20}\end{matrix}\right.< =>\left[{}\begin{matrix}x=\left(4+2\sqrt{5}\right)^2=36+16\sqrt{5}\left(TMDK\right)\\x=\left(4-2\sqrt{5}\right)^2=36-16\sqrt{5}\left(TMDK\right)\end{matrix}\right.\)
