C=\(\left(\frac{2+\sqrt{a}}{2-\sqrt{a}}-\frac{2-\sqrt{a}}{2+\sqrt{a}}-\frac{4a}{a-4}\right):\left(\frac{2}{2-\sqrt{a}}-\frac{\sqrt{a}+3}{2\sqrt{a}-a}\right)\)
\(\Leftrightarrow C=\left(\frac{\left(2+\sqrt{a}\right)^2-\left(2-\sqrt{a}\right)^2+4a}{4-a}\right)\):\(\left(\frac{2\sqrt{a}-\sqrt{a}-3}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)
\(\Leftrightarrow C=\frac{4+4\sqrt{a}+a-4+4\sqrt{a}-a+4a}{4-a}\).\(\frac{\sqrt{a}\left(2-\sqrt{a}\right)}{\sqrt{a}-3}\)
\(\Leftrightarrow C=\frac{4\sqrt{a}(\sqrt{a}+2)}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}.\frac{\sqrt{a}\left(2-\sqrt{a}\right)}{\sqrt{a}-3}\)
\(\Leftrightarrow C=\frac{4a}{\sqrt{a}-3}\)
b. Để C>0 thì \(\sqrt{a}\)-3>0 ( Do 4\(\sqrt{a}\)>0 với mọi a>0)
\(\Leftrightarrow\sqrt{a}>3\Leftrightarrow\text{}a>9\)
Vậy khi a>9 thì C>0
c. C=-1
\(\Leftrightarrow\) \(\frac{4a}{\sqrt{a}-3}=-1\Leftrightarrow4a=3-\sqrt{a}\Leftrightarrow4a+\sqrt{a}-3=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=-1\:\left(loai\right)\\\sqrt{a}=\frac{3}{4}\end{matrix}\right.\: \Leftrightarrow a=\frac{9}{16}\)
Vậy khi a=9/16 thì C=-1