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a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
b. \(Q=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
c. Để \(Q< 1\Rightarrow Q-1< 0\Leftrightarrow\frac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\Leftrightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)
Vậy \(0\le x< 9\)thì \(Q< 1\)
\(A=\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right):\left(\frac{1}{x-\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}-1}\right)\) Đkxđ : x > 1
\(A=\left(\frac{\sqrt{x}-1}{x-1}+\frac{2}{x-1}\right):\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(A=\frac{\sqrt{x}-1+2}{x-1}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{1-x}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(1-x\right)}\)
\(A=\frac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(1-x\right)}=\frac{\sqrt{x}}{1-x}\)