a)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{x-1}\right):\frac{1}{\sqrt{x}+1}\)
\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{1}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+1-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}+1\right)\)
\(A=\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}+1\right)\)
\(A=\frac{1}{\sqrt{x}-1}\)
b)với \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)để A\(< -\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{\sqrt{x}-1}< -\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{\sqrt{x}-1}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{2+\sqrt{x}-1}{2\left(\sqrt{x}-1\right)}< 0\)
\(\Leftrightarrow\frac{1+\sqrt{x}}{2\left(\sqrt{x}-1\right)}< 0\)
vì 1+\(\sqrt{x}\)>0 nên 2(\(\sqrt{x}-1\))<0
\(\Leftrightarrow\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
\(\Leftrightarrow x>1\)
