a: \(P=\left(\dfrac{3\sqrt{a}}{\sqrt{a}+4}+\dfrac{\sqrt{a}}{\sqrt{a}-4}+\dfrac{4\left(a+2\right)}{16-a}\right):\left(1-\dfrac{2\sqrt{a}+5}{\sqrt{a}+4}\right)\)
\(=\left(\dfrac{3\sqrt{a}}{\sqrt{a}+4}+\dfrac{\sqrt{a}}{\sqrt{a}-4}-\dfrac{4\left(a+2\right)}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}\right):\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\sqrt{a}+4}\)
\(=\dfrac{3\sqrt{a}\left(\sqrt{a}-4\right)+\sqrt{a}\left(\sqrt{a}+4\right)-4a-8}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}\cdot\dfrac{\sqrt{a}+4}{-\sqrt{a}-1}\)
\(=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}-4\right)\left(-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(=\dfrac{-8\sqrt{a}-8}{\left(-\sqrt{a}-1\right)\left(\sqrt{a}-4\right)}=\dfrac{8}{\sqrt{a}-4}\)
b: Để P=-3 thì \(\sqrt{a}-4=\dfrac{8}{-3}=-\dfrac{8}{3}\)
=>\(\sqrt{a}=-\dfrac{8}{3}+4=-\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{4}{3}\)
=>\(a=\dfrac{16}{9}\left(nhận\right)\)
c: Để P là số nguyên tố thì \(8⋮\sqrt{a}-4\)
=>\(\sqrt{a}-4\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\) và \(P\in p\)
=>\(\left\{{}\begin{matrix}\sqrt{a}\in\left\{5;3;6;2;8;0;12;-4\right\}\\\sqrt{a}-4>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a\in\left\{25;9;36;4;64;0;144\right\}\\a>16\end{matrix}\right.\)
=>\(a\in\left\{25;36;64;144\right\}\)
Thay lại vào P, ta thấy a=64 thỏa mãn
vậy: a=64
