a: \(N=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\left(\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\right)\)
\(=\dfrac{2x+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}-2-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để N nguyên thì \(\sqrt{x}-1+2⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{4;9\right\}\)
c: Để N=cănN thì N=1 hoặc N=0
=>N=1
=>\(\sqrt{x}+2=\sqrt{x}-2\)
=>2=-2(loại)
