a: \(A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b: Để A là số nguyên thì \(2\sqrt{x}+2-3⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;3\right\}\)
=>x=0 hoặc x=4
c: \(A-1=\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
=>A<1