Bài 2:
ĐKXĐ: x>=0 và x<>4
a: \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{2+5\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b: Để P là số nguyên thì \(3\sqrt{x}⋮\sqrt{x}+2\)
=>\(3\sqrt{x}+6-6⋮\sqrt{x}+2\)
=>\(-6⋮\sqrt{x}+2\)
=>\(\sqrt{x}+2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(\sqrt{x}\in\left\{-1;-3;0;-4;1;-5;4;-8\right\}\)
=>\(\sqrt{x}\in\left\{0;1;4\right\}\)
=>\(x\in\left\{0;1;16\right\}\)
