Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
x = 2\left( {tmdk} \right)\\
A = \dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{2 + \sqrt 2 + 1}}{{\sqrt 2 + 1}}\\
= \dfrac{{\left( {3 + \sqrt 2 } \right)\left( {\sqrt 2 - 1} \right)}}{{2 - 1}}\\
= 3\sqrt 2 - 3 + 2 - \sqrt 2 \\
= 2\sqrt 2 - 1\\
b)B = \dfrac{1}{{\sqrt x - 1}} - \dfrac{{x + 2}}{{x\sqrt x - 1}} - \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x - 1}} - \dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\\
= \dfrac{{x + \sqrt x + 1 - x - 2 - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1 - x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - \sqrt x }}{{x + \sqrt x + 1}}\\
c)C = - A.B\\
= - \dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}}.\dfrac{{ - \sqrt x }}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 1}}{{\sqrt x + 1}}\\
= 1 - \dfrac{1}{{\sqrt x + 1}}\\
C \in Z\\
\Rightarrow \dfrac{1}{{\sqrt x + 1}} \in Z\\
\Rightarrow \sqrt x + 1 = 1\\
\Rightarrow \sqrt x = 0\\
\Rightarrow x = 0
\end{array}$
Vậy x=0 thì C nguyên.
