Chắc bạn ghi nhầm đề \(x=3-2\sqrt{3}\) thì \(x< 0\) nên \(\sqrt{x}\) không xác định, đề đúng phải là \(x=3-2\sqrt{2}\)
ĐKXĐ: \(x>0;x\ne1\)
\(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\)
\(\Rightarrow Q=\frac{\sqrt{2}-1-1}{\sqrt{2}-1+1}=\frac{\sqrt{2}-2}{\sqrt{2}}=1-\sqrt{2}\)
\(A=\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{4}{\sqrt{x}}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}+\frac{x-\sqrt{x}+1}{\sqrt{x}}-\frac{4}{\sqrt{x}}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x+\sqrt{x}+1+x-\sqrt{x}+1-4}{\sqrt{x}}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\frac{2\left(x-1\right)}{\sqrt{x}}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Để \(A\sqrt{x}< 8\Leftrightarrow\frac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}.\sqrt{x}< 8\)
\(\Rightarrow\left(\sqrt{x}-1\right)^2< 4\Rightarrow\sqrt{x}-1< 2\Rightarrow\sqrt{x}< 3\)\(\Rightarrow x< 9\)
Kết hợp ĐKXĐ \(\Rightarrow\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)
