Ta có: \(B=\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}+\frac{5-2\sqrt{x}}{x+\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\frac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-4-\sqrt{x}+1+5-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
Để \(\frac{A}{B}< 4\) thì \(\frac{A}{B}-4< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}-2}{\sqrt{x}+2}-4< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}-\frac{4\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{4x+8\sqrt{x}-4\left(x-7\sqrt{x}+10\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{4x+8\sqrt{x}-4x+28\sqrt{x}-40}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{36\sqrt{x}-40}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}< 0\)
Trường hợp 1:
\(\left\{{}\begin{matrix}36\sqrt{x}-40< 0\\\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}36\sqrt{x}< 40\\\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-5>0\\\sqrt{x}-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}-2< 0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}< \frac{10}{9}\\\left[{}\begin{matrix}\sqrt{x}>5\\\sqrt{x}< 2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\left(loại\right)\\\left[{}\begin{matrix}x>25\\x< 4\end{matrix}\right.\end{matrix}\right.\)
=> Loại
Trường hợp 2:
