Đáp án:
$\begin{array}{l}
a)\sqrt {x - 1} + \frac{3}{{x - 1}} > \sqrt {1 - x} \\
Dk:\left\{ \begin{array}{l}
x - 1 \ge 0\\
x \ne 1\\
1 - x \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ge 1\\
x \le 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 1\\
x = 1
\end{array} \right.\\
\Rightarrow Ko\,có\,gtri\,của\,x\,để\,bpt\,xác\,định\\
b)\\
x + 2 - \frac{1}{{\sqrt {x + 2} }} \ge x + 1\\
Dkxd:\left\{ {x + 2 > 0} \right. \Rightarrow x > - 2
\end{array}$
