Giải thích các bước giải:
Bài 1:
Các biểu thức đã cho có nghĩa khi và chỉ khi:
\(\begin{array}{l}
\sqrt { - 2x + 3} :\,\,\,\,\, - 2x + 3 \ge 0 \Leftrightarrow 2x - 3 \le 0 \Leftrightarrow x \le \frac{3}{2}\\
\sqrt { - 5x} :\,\,\,\,\,\,\,\,\, - 5x \ge 0 \Leftrightarrow 5x \le 0 \Leftrightarrow x \le 0\\
\sqrt {\frac{x}{3}} :\,\,\,\,\,\,\,\,\,\,\,\,\frac{x}{3} \ge 0 \Leftrightarrow x \ge 0\\
\sqrt {\frac{1}{{ - 1 + x}}} :\,\,\,\,\,\,\,\,\, - 1 + x > 0 \Leftrightarrow x - 1 > 0 \Leftrightarrow x > 1\\
\sqrt {{x^2} - 2x + 1} :\,\,\,\,\,\,\,\,\,\,\,\,{x^2} - 2x + 1 \ge 0 \Leftrightarrow {\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {\frac{2}{{{x^2}}}} :\,\,\,\,\,\,\,\,\,\,x \ne 0\\
\sqrt { - {x^2} - 2x - 1} :\,\,\,\,\,\,\,\,\, - {x^2} - 2x - 1 \ge 0 \Leftrightarrow {x^2} + 2x + 1 \le 0 \Leftrightarrow {\left( {x + 1} \right)^2} \le 0 \Leftrightarrow x = - 1\\
\sqrt {\frac{{x - 1}}{{x + 2}}} :\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}
\frac{{x - 1}}{{x + 2}} \ge 0\\
x + 2 \ne 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x < - 2
\end{array} \right.
\end{array}\)
Bài 2:
\(\begin{array}{l}
a,\\
2 = 1 + 1 = \sqrt 1 + 1 < \sqrt 2 + 1\\
b,\\
1 = 2 - 1 = \sqrt 4 - 1 > \sqrt 3 - 1\\
c,\\
2\sqrt {31} > 2\sqrt {25} = 2.5 = 10\\
d,\\
3\sqrt {11} < 3.\sqrt {16} = 3.4 = 12\\
\Rightarrow - 3\sqrt {11} > - 12
\end{array}\)
