\(S=\frac{2\sqrt{x}}{x-2\sqrt{x}}+\frac{3\sqrt{x}-14}{x-4}\)
a, \(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow S=\frac{2}{\sqrt{x}-2}+\frac{3\sqrt{x}-14}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow S=\frac{2\sqrt{x}+4+3\sqrt{x}-14}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow S=\frac{5\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{5}{\sqrt{x}+2}\)
b, Để S nhận giá trị nguyên
\(\Rightarrow\sqrt{x}+2\inƯ_{\left(5\right)}=\left\{1;-1;5;-5\right\}\)
mà x \(\ge\) 0 \(\Rightarrow\sqrt{x}+2\ge0\) \(\Rightarrow\) \(\left[{}\begin{matrix}\sqrt{x}+2=1\\\sqrt{x}+2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(L\right)\\\sqrt{x}=3\end{matrix}\right.\)
\(\Rightarrow x=9\)
Vậy x = 9 thì S nhận giá trị nguyên
