Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 3A.\\ a.\ A=\frac{1}{\sqrt{x} -1}\\ b.\ B=\frac{\sqrt{x} +1}{\sqrt{x}}\\ c.\ C=\frac{x-1}{\sqrt{x}}\\ 3B.\\ a.\ A=\frac{\sqrt{x} -3}{2}\\ b.\ B=\frac{x+9}{x-9}\\ c.\ C=\frac{x+9}{2\left(\sqrt{x} +3\right)} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 3A.\\ a.\ A=\frac{\sqrt{x} -1+2}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)} =\frac{\sqrt{x} +1}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)} =\frac{1}{\sqrt{x} -1}\\ b.\ B=\frac{x-1}{\sqrt{x}\left(\sqrt{x} -1\right)} =\frac{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)}{\sqrt{x}\left(\sqrt{x} -1\right)} =\frac{\sqrt{x} +1}{\sqrt{x}}\\ c.\ C=\frac{\sqrt{x} +1}{\sqrt{x}} :\frac{1}{\sqrt{x} -1} =\frac{\sqrt{x} +1}{\sqrt{x}} .\left(\sqrt{x} -1\right) =\frac{x-1}{\sqrt{x}}\\ 3B.\\ a.\ A=\frac{\left(\sqrt{x} -3\right)\left(\sqrt{x} +3\right)}{2\left(\sqrt{x} +3\right)} =\frac{\sqrt{x} -3}{2}\\ b.\ B=\frac{\sqrt{x}\left(\sqrt{x} +3\right) -3\left(\sqrt{x} -3\right)}{\left(\sqrt{x} +3\right)\left(\sqrt{x} -3\right)} =\frac{x+3\sqrt{x} -3\sqrt{x} +9}{\left(\sqrt{x} +3\right)\left(\sqrt{x} -3\right)} =\frac{x+9}{x-9}\\ c.\ C=\frac{x+9}{x-9} .\frac{\sqrt{x} -3}{2} =\frac{x+9}{\left(\sqrt{x} +3\right)\left(\sqrt{x} -3\right)} .\frac{\sqrt{x} -3}{2} =\frac{x+9}{2\left(\sqrt{x} +3\right)} \end{array}$
