Đáp án:
$\begin{array}{l}
1)\sqrt {{x^2} - 1} \ge \sqrt {{x^2} - 2x + 1} \\
Đkxđ:\left\{ \begin{array}{l}
{x^2} - 1 \ge 0\\
{x^2} - 2x + 1 \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x^2} \ge 1\\
{\left( {x - 1} \right)^2} \ge 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le - 1
\end{array} \right.\\
\Rightarrow {x^2} - 1 \ge {x^2} - 2x + 1\\
\Rightarrow 2x \ge 2\\
\Rightarrow x \ge 1\\
Vậy\,x \ge 1\\
2)Dkxd:x \ge 5\\
\sqrt {x - 2} \ge \sqrt {x - 5} \\
\Rightarrow x - 2 \ge x - 5\\
\Rightarrow - 2 \ge - 5\left( {luon\,dung} \right)\\
Vậy\,x \ge 5\\
3)Dkxd:\left\{ \begin{array}{l}
x + \frac{3}{2} \ge 0\\
x + \frac{4}{3} \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge - \frac{3}{2}\\
x \ge - \frac{4}{3}
\end{array} \right. \Rightarrow x \ge - \frac{3}{2}\\
\sqrt {x + \frac{3}{2}} < x + \frac{4}{3}\\
\Rightarrow x + \frac{3}{2} < {x^2} + \frac{8}{3}x + \frac{{16}}{9}\\
\Rightarrow {x^2} + \frac{5}{3}x + \frac{5}{{18}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > \frac{{ - 5 + \sqrt {15} }}{6}\\
x < \frac{{ - 5 - \sqrt {15} }}{6}
\end{array} \right.\\
\Rightarrow x > \frac{{ - 5 + \sqrt {15} }}{6}
\end{array}$
