a: ĐKXĐ: x>=0; x<>4
\(Q=\frac{2}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{2\sqrt{x}}{x-4}\)
\(=\frac{2}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{2\left(\sqrt{x}-2\right)-\sqrt{x}-2+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2\sqrt{x}-4+\sqrt{x}-2}{\left.\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\right.}\)
\(=\frac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{3}{\sqrt{x}+2}\)
b: \(Q=\frac65\)
=>\(\frac{3}{\sqrt{x}+2}=\frac65\)
=>\(\sqrt{x}+2=3\cdot\frac56=\frac52\)
=>\(\sqrt{x}=\frac52-2=\frac12\)
=>\(x=\frac14\) (nhận)
c: Để Q nguyên thì 3⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2=3\)
=>\(\sqrt{x}=1\)
=>x=1(nhận)
