ĐKXĐ: \(a>0\) ; \(a\ne9\)
\(A=\left(\frac{\sqrt{a}\left(\sqrt{a}+3\right)+\sqrt{a}\left(\sqrt{a}-3\right)}{a-9}\right).\frac{\left(a-9\right)}{\sqrt{a}}\)
\(A=\frac{\sqrt{a}\left(2\sqrt{a}\right)}{\left(a-9\right)}.\frac{\left(a-9\right)}{\sqrt{a}}=2\sqrt{a}\)
Để \(A=3\sqrt{a}-16\)
\(\Leftrightarrow2\sqrt{a}=3\sqrt{a}-16\)
\(\Rightarrow\sqrt{a}=16\)
\(\Rightarrow a=16^2=256\)