Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ C=\frac{2}{\sqrt{x} -2}\\ b.\ x=64\\ c.\ x >1;\ x\neq 4 \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ C=\frac{\sqrt{x} -2+\sqrt{x} +2}{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)} .\frac{x+4\sqrt{x} +4}{\sqrt{x}\left(\sqrt{x} +2\right)}\\ C=\frac{2\sqrt{x}}{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)} .\frac{\left(\sqrt{x} +2\right)^{2}}{\sqrt{x}\left(\sqrt{x} +2\right)}\\ C=\frac{2}{\sqrt{x} -2}\\ b.\ C=\frac{1}{3}\\ \Leftrightarrow \frac{2}{\sqrt{x} -2} =\frac{1}{3}\\ \Leftrightarrow \sqrt{x} -2=6\\ \Leftrightarrow x=64\ ( TM)\\ c.\ C >-2\\ \Leftrightarrow C+2 >0\\ \Leftrightarrow \frac{2}{\sqrt{x} -2} +2 >0\\ \Leftrightarrow \frac{2\sqrt{x} -2}{\sqrt{x} -2} >0\\ \Leftrightarrow 2\sqrt{x} -2 >0\ \left( do\ \sqrt{x} +2 >0\right)\\ \Leftrightarrow \sqrt{x} >1\\ \Leftrightarrow x >1\ kết\ hợp\ ĐK\ \\ \Rightarrow x >1;\ x\neq 4 \end{array}$
