\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x-2}}\right).\frac{\sqrt{x}-2}{\sqrt{x}}\)
a, \(Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x-2}}\right).\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}+2}.\frac{1}{\sqrt{x}}\)
\(=\frac{2}{\sqrt{x}+2}\)
\(b,x>0;x\ne4\)
\(A>\frac{1}{2}\)
\(\Rightarrow\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)
\(\Rightarrow4>\sqrt{x}+2\)
\(\Rightarrow\sqrt{x}< 2\)
\(\Rightarrow x< 4\)
Vậy \(0< x< 4\)
c, \(B=\frac{5}{2}.A=\frac{5}{2}.\frac{2}{\sqrt{x}+2}=\frac{5}{\sqrt{x}+2}\)
\(B\in Z\)
\(\Rightarrow\left(\sqrt{x}+2\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Vì: \(\sqrt{x}+2>2\forall x>0;x\ne4\)
\(\Rightarrow\sqrt{x}+2=5\)
\(\Rightarrow\sqrt{x}=3\)
\(\Rightarrow x=9\left(tm\right)\)
Vậy ...................................................
