Bài 1. a. Ta có:
\(\eqalign{ & A = {{\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} } \over {\sqrt 2 \left( {\sqrt 5 - \sqrt 3 } \right)}} \cr&= {{\left| {\sqrt 5 - \sqrt 3 } \right|} \over {\sqrt 2 \left( {\sqrt 5 - \sqrt 3 } \right)}} \cr & \,\,\,\,\, = {{\sqrt 5 - \sqrt 3 } \over {\sqrt 2 \left( {\sqrt 5 - \sqrt 3 } \right)}} = {1 \over {\sqrt 2 }} \cr} \)
b. Ta có:
\(\eqalign{ & B = {1 \over {a{b^2}}}.{{\sqrt {{a^2}{b^4}} } \over {\sqrt 3 }} = {{\left| a \right|.{b^2}} \over {\sqrt 3 a{b^2}}} \cr & \,\,\,\,\, = \left\{ {\matrix{ {{1 \over {\sqrt 3 }}\,\text{ nếu }\,a > 0;b \ne 0} \cr {{{ - 1} \over {\sqrt 3 }}\,\text{ nếu }\,a < 0;b \ne 0.} \cr } } \right. \cr} \)
Bài 2. Ta có: \({{\sqrt {x + 1} } \over {\sqrt {x - 1} }} = 2 \Leftrightarrow \left\{ {\matrix{ {x > 1} \cr {\sqrt {{{x + 1} \over {x - 1}}} = 2} \cr } } \right. \)
\(\Leftrightarrow \left\{ {\matrix{ {x > 1} \cr {{{x + 1} \over {x - 1}} = 4} \cr } } \right. \Leftrightarrow \left\{ {\matrix{ {x > 1} \cr {x = {5 \over 3}} \cr } } \right.\)
Bài 3. Ta có:
\(\eqalign{ & \sqrt {{{ - 1} \over {x - 1}}} < 1 \Leftrightarrow \sqrt {{1 \over {1 - x}}} < 1 \cr&\Leftrightarrow \left\{ {\matrix{ {1 - x > 0} \cr {{1 \over {1 - x}} < 1} \cr } } \right. \cr & \Leftrightarrow \left\{ {\matrix{ {x < 1} \cr {1 < 1 - x} \cr } } \right. \Leftrightarrow x < 0 \cr} \)
