1. ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)
2.
\(A=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\\ =\frac{\sqrt{a}-2}{3\sqrt{a}}\)
3. Ta thấy \(\sqrt{a}>0\forall a>0\)(ĐKXĐ)
Nên để A > 0 thì \(\left\{{}\begin{matrix}\sqrt{a}-2>0\\3\sqrt{a}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>4\\a>0\end{matrix}\right.\Leftrightarrow a>4\)
ĐKXĐ \(a\ne1;a\ne4;a>0\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)=\frac{\sqrt{a}-2}{\sqrt{a}}\)
Có \(A=\frac{\sqrt{a}-2}{\sqrt{a}}\)
Vậy ta có để \(A>0\Leftrightarrow\frac{\sqrt{a}-2}{\sqrt{a}}>0\)
Còn lại chắc bạn làm đc mà