Sửa đề: \(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
a) Ta có: \(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{4-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3+4-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Để \(A< \frac{1}{2}\) thì \(A-\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\)
mà \(2\left(\sqrt{x}+1\right)>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-3< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow\left|x\right|< 9\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x< 3\end{matrix}\right.\)
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}0\le x< 3\\x\ne1\end{matrix}\right.\)
Vậy: Để \(A< \frac{1}{2}\) thì \(\left\{{}\begin{matrix}0\le x< 3\\x\ne1\end{matrix}\right.\)
