Đáp án:
$\begin{array}{l}
B1)Dkxd:x \ge 0;x \ne 1\\
a)C = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{x\sqrt x - x + \sqrt x - 1}}} \right):\left( {\dfrac{{\sqrt x }}{{x + 1}} - 1} \right)\\
= \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}} \right):\dfrac{{\sqrt x - x - 1}}{{x + 1}}\\
= \dfrac{{x + 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{x + 1}}{{\sqrt x - x - 1}}\\
= \dfrac{{x - 1}}{{\left( {\sqrt x - 1} \right)\left( { - x + \sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{ - x + \sqrt x - 1}}\\
b)Khi:x \ge 0;x \ne 1 \Rightarrow \sqrt x + 1 > 0\\
- x + \sqrt x - 1\\
= - \left( {x - \sqrt x } \right) - 1\\
= - \left( {x - 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{1}{4} - 1\\
= - {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} - \dfrac{3}{4} \le - \dfrac{3}{4} < 0\\
\Rightarrow \dfrac{{\sqrt x + 1}}{{ - x + \sqrt x - 1}} < 0\left( {khi:x \ge 0;x \ne 1} \right)\\
\Rightarrow C < 0
\end{array}$
Vậy C luôn nhận giá trị âm với mọi x thích hợp
Bài 2:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
A = \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x - 1}}} \right).\left( {\dfrac{{x - 1}}{{\sqrt x - 1}} - 2} \right)\\
= \dfrac{2}{{\sqrt x - 1}}.\left( {\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}} - 2} \right)\\
= \dfrac{2}{{\sqrt x - 1}}.\left( {\sqrt x + 1 - 2} \right)\\
= \dfrac{2}{{\sqrt x - 1}}.\left( {\sqrt x - 1} \right)\\
= 2
\end{array}$
b) A luôn có giá trị nguyên là 2 với mọi x nguyên thỏa mãn $x \ge 0;x \ne 1$
