1: Thay x=16 vào A, ta được:
\(A=\dfrac{16-2\cdot4}{4+3}=\dfrac{16-8}{7}=\dfrac{8}{7}\)
2: \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{3\sqrt{x}+6}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-2\cdot\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{3\left(\sqrt{x}+2\right)}{x+4}\)
\(=\dfrac{x+2\sqrt{x}-2\sqrt{x}+4}{\sqrt{x}-2}\cdot\dfrac{3}{x+4}=\dfrac{3}{\sqrt{x}-2}\)
3: Đặt P=A*B
\(=\dfrac{3}{\sqrt{x}-2}\cdot\dfrac{x-2\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\)
Để P là số nguyên thì \(3\sqrt{x}⋮\sqrt{x}+3\)
=>\(3\sqrt{x}+9-9⋮\sqrt{x}+3\)
=>\(-9⋮\sqrt{x}+3\)
=>\(\sqrt{x}+3\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(\sqrt{x}+3\in\left\{3;9\right\}\)
=>\(\sqrt{x}\in\left\{0;6\right\}\)
=>\(x\in\left\{0;36\right\}\)
giải giúp a b 
