Bài 1. a. \(\sqrt {2x - 3} \) có nghĩa \( \Leftrightarrow 2x - 3 \ge 0 \Leftrightarrow x \ge {3 \over 2}\)
b. \(\sqrt {{1 \over {2 - x}}} \) có nghĩa \( \Leftrightarrow {1 \over {2 - x}} \ge 0 \Leftrightarrow 2 - x > 0 \Leftrightarrow x < 2\)
c. \(\sqrt {x + 1} + \sqrt {1 - x} \) có nghĩa \( \Leftrightarrow \left\{ {\matrix{ {x + 1 \ge 0} \cr {1 - x \ge 0} \cr } } \right. \Leftrightarrow \left\{ {\matrix{ {x \ge - 1} \cr {x \le 1} \cr } } \right.\)
\( \Leftrightarrow - 1 \le x \le 1\)
Bài 2.
a. Ta có:
\(\eqalign{ & \sqrt {9 - 4\sqrt 5 } - \sqrt 5 \cr& = \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - \sqrt 5 \cr & = \left| {\sqrt 5 - 2} \right| - \sqrt 5 \cr & = \sqrt 5 - 2 - \sqrt 5 = - 2 \cr} \)
\(( {\text{Vì}\,\sqrt 5 - 2 > 0 \Rightarrow \left| {\sqrt 5 - 2} \right| = \sqrt 5 - 2} ) \)
b. Ta có:
\(\eqalign{ & \sqrt {3 - 2\sqrt 2 } - \sqrt {3 + 2\sqrt 2 } \cr&= \sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} \cr & = \left| {1 - \sqrt 2 } \right| - \left| {1 + \sqrt 2 } \right| \cr&= - \left( {1 - \sqrt 2 } \right) - \left( {1 + \sqrt 2 } \right) = - 2 \cr} \)
\(( \text{Vì}\,1 - \sqrt 2 < 0 \) \(\Rightarrow \left| {1 - \sqrt 2 } \right| = - \left( {1 - \sqrt 2 } \right) )\)
