Bài 3:
a: Ta có: \(Q=\dfrac{2}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{x-4}\)
\(=\dfrac{2\sqrt{x}-4-\sqrt{x}-2+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+2}\)
b: Để \(Q=\dfrac{6}{5}\) thì \(\sqrt{x}+2=\dfrac{5}{2}\)
hay \(x=\dfrac{1}{4}\)
c: Để Q nguyên thì \(3⋮\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}+2=3\)
hay x=1
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