ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+2\ne0\\\sqrt{x}-2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có : \(\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\frac{1}{\sqrt{x}-2}\)
\(=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{7}{x-4}\right)\left(\sqrt{x}-2\right)\)
\(=\frac{\left(\sqrt{x}-2+7\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
<=>\((\frac{1}{\sqrt{x}+2}+\frac{7}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}):\frac{1}{\sqrt{x}-2}\)
<=> \((\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}+\frac{7}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}):\frac{1}{\sqrt{x}-2}\)
<=>\(\frac{\sqrt{x}+5}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}.\sqrt{x}-2\)
<=>\(\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
Vậy biểu thức sau khi rút gọn có dạng \(\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
