\(B=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right)\)
\(=\frac{\left(2\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)
\(C=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)}\left(\frac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)}.\frac{-2}{\left(\sqrt{x}+2\right)}=\frac{-2}{\sqrt{x}+2}\)
Để C nguyên \(\Rightarrow\sqrt{x}+2=Ư\left(2\right)=2\)
\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
a/ ĐKXĐ:...
\(B=\left(\frac{\sqrt{x}+\sqrt{x}+2}{x-4}\right).\frac{x-4}{\sqrt{x}+2}=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)
\(C=\frac{\sqrt{x}+2}{\sqrt{x}-2}\left(\frac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)
\(C=\frac{\sqrt{x}+2}{\sqrt{x}-2}.\frac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}=\frac{-2}{\sqrt{x}-2}\)
Để C đạt GT nguyên
\(\Leftrightarrow\sqrt{x}-2\inƯ_{\left(-2\right)}=\left\{\pm2;\pm1\right\}\)
\(\left[{}\begin{matrix}x=0\\x=9\\x=1\\x=16\end{matrix}\right.\)