Bài 1. Rút gọn :
a. \(A = {{\sqrt {8 – 2\sqrt {15} } } \over {\sqrt {10} – \sqrt 6 }}\)
b. \(B = {1 \over {a{b^2}}}.\sqrt {{{{a^2}{b^4}} \over 3}} \)
Bài 2. Tìm x, biết : \({{\sqrt {x + 1} } \over {\sqrt {x – 1} }} = 2\)
Bài 3. Tìm x, biết : \(\sqrt {{{ – 1} \over {x – 1}}} < 1\)
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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:
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\(\eqalign{ & B = {1 \over {a{b^2}}}.{{\sqrt {{a^2}{b^4}} } \over {\sqrt 3 }} = {{\left| a \right|.{b^2}} \over {\sqrt 3 a{b^2}}} \cr & \,\,\,\,\, = \left\{ {\matrix{ {{1 \over {\sqrt 3 }}\,\text{ nếu }\,a > 0;b \ne 0} \cr {{{ – 1} \over {\sqrt 3 }}\,\text{ nếu }\,a < 0;b \ne 0.} \cr } } \right. \cr} \)
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:
\(\eqalign{ & \sqrt {{{ – 1} \over {x – 1}}} < 1 \Leftrightarrow \sqrt {{1 \over {1 – x}}} < 1 \cr&\Leftrightarrow \left\{ {\matrix{ {1 – x > 0} \cr {{1 \over {1 – x}} < 1} \cr } } \right. \cr & \Leftrightarrow \left\{ {\matrix{ {x < 1} \cr {1 < 1 – x} \cr } } \right. \Leftrightarrow x < 0 \cr} \)
