ĐKXĐ: \(a>0;a\ne4;9\)
\(C=\left(\frac{\left(2+\sqrt{a}\right)^2}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}-\frac{\left(2-\sqrt{a}\right)^2}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}+\frac{4a}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\right):\left(\frac{2\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}-\frac{\sqrt{a}+3}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)
\(=\left(\frac{a+4\sqrt{a}+4-a+4\sqrt{a}-4+4a}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\right):\left(\frac{2\sqrt{a}-\sqrt{a}-3}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)
\(=\frac{4\sqrt{a}\left(\sqrt{a}+2\right)}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}.\frac{\sqrt{a}\left(2-\sqrt{a}\right)}{\left(\sqrt{a}-3\right)}=\frac{4a}{\sqrt{a}-3}\)
\(C=-1\Leftrightarrow\frac{4a}{\sqrt{a}-3}=-1\)
\(\Leftrightarrow4a+\sqrt{a}-3=0\Leftrightarrow\left(\sqrt{a}+1\right)\left(4\sqrt{a}-3\right)=0\)
\(\Leftrightarrow4\sqrt{a}-3=0\Leftrightarrow\sqrt{a}=\frac{3}{4}\Rightarrow a=\frac{9}{16}\)
