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