\(1,\\ a,A=2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\\ b,B=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{5}-2=3-\sqrt{5}+\sqrt{5}-2=1\\ 2,\\ a,x=36\Leftrightarrow\sqrt{x}=6\Leftrightarrow A=\dfrac{6+4}{6+2}=\dfrac{10}{8}=\dfrac{5}{4}\\ b,B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\dfrac{\sqrt{x}+2}{x+16}\\ B=\dfrac{\left(x+16\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)\left(x+16\right)}=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\)
\(c,P=B\left(A-1\right)=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\\ P=\dfrac{1}{\sqrt{x}-4}\in Z\Leftrightarrow1⋮\sqrt{x}-4\\ \Leftrightarrow\sqrt{x}-4\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{3;5\right\}\\ \Leftrightarrow x\in\left\{9;25\right\}\)
