a: \(A=\sqrt{x}+2=5+2=7\)
b: \(B=\dfrac{2\sqrt{x}+4+3\sqrt{x}-6+x-5\sqrt{x}+2}{x-4}\)
\(=\dfrac{x}{x-4}\)
c: \(P=A\cdot B=\dfrac{x}{x-4}\cdot\dfrac{x-4}{\sqrt{x}-2}=\dfrac{x}{\sqrt{x}-2}\)
Để P là số nguyên thì \(x-4+4⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{9;1;16;0;36\right\}\)