Bài 2:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)
\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{2+5\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5x-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b: Để P là số nguyên thì \(3\sqrt{x}⋮\sqrt{x}+2\)
=>\(3\sqrt{x}+6-6⋮\sqrt{x}+2\)
=>\(-6⋮\sqrt{x}+2\)
mà \(\sqrt{x}+2>=2\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}+2\in\left\{2;3;6\right\}\)
=>\(\sqrt{x}\in\left\{0;1;4\right\}\)
=>\(x\in\left\{0;1;16\right\}\)
