Bài 1. a. A có nghĩa \( \Leftrightarrow \left\{ {\matrix{ {x - 3 \ne 0} \cr {x - 3 \ge 0} \cr } } \right. \Leftrightarrow x - 3 > 0 \Leftrightarrow x > 3\)
b. B có nghĩa \( \Leftrightarrow \left\{ {\matrix{ {x - 2 \ge 0} \cr {x - 2 \ne 0} \cr } } \right. \Leftrightarrow \left\{ {\matrix{ {x \ge 2} \cr {x \ne 2} \cr } } \right. \Leftrightarrow x > 2\)
Bài 2. a. Ta có:
\(\eqalign{ & 2\sqrt {2 + \sqrt 3 } = \sqrt {4\left( {2 + \sqrt 3 } \right)} \cr & = \sqrt {8 + 4\sqrt 3 } = \sqrt {6 + 2\sqrt {12} + 2} \cr & = \sqrt {{{\left( {\sqrt 6 + \sqrt 2 } \right)}^2}} = \left| {\sqrt 6 + \sqrt 2 } \right| \cr & = \sqrt 2 + \sqrt 6 \,\,\left( {đpcm} \right) \cr} \)
b. Ta có:
\(\eqalign{ & \sqrt {1 + {{\sqrt 3 } \over 2}} = \sqrt {{{2 + \sqrt 3 } \over 2}} \cr & = \sqrt {{{4 + 2\sqrt 3 } \over 4}} = {{\sqrt {{{\left( {1 + \sqrt 3 } \right)}^2}} } \over {\sqrt 4 }} \cr & = {{1 + \sqrt 3 } \over 2}\,\,\left( {đpcm} \right) \cr} \)
Bài 3. a. Ta có:
\(\eqalign{ A &= \left( {\sqrt {21} + 3} \right)\sqrt {10 - 2\sqrt {21} } \cr & = \sqrt 3 \left( {\sqrt 7 + \sqrt 3 } \right)\sqrt {{{\left( {\sqrt 7 - \sqrt 3 } \right)}^2}} \cr & = \sqrt 3 .\left( {\sqrt 7 + \sqrt 3 } \right)\left( {\sqrt 7 - \sqrt 3 } \right) \cr&= 4\sqrt 3 \cr} \)
b. Ta có:
\(\eqalign{ B& = \left( {\sqrt 5 - 1} \right)\sqrt {6 + 2\sqrt 5 } \cr & = \left( {\sqrt 5 - 1} \right)\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} \cr & = \left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right) \cr & = 5 - 1 = 4 \cr} \)
Bài 4. a. Ta có:
\(\eqalign{ & P = \left[ {{1 \over {\sqrt x + 1}} - {1 \over {\sqrt x \left( {\sqrt x + 1} \right)}}} \right]:{{x - \sqrt x + 1} \over {{{\left( {\sqrt x } \right)}^3} + 1}} \cr & = {{\sqrt x - 1} \over {\sqrt x \left( {\sqrt x + 1} \right)}}:{{x - \sqrt x + 1} \over {\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} \cr & = {{\sqrt x - 1} \over {\sqrt x \left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right) \cr&= {{\sqrt x - 1} \over {\sqrt x }} \cr} \)
b. Ta có: \(P < 0\) (điều kiện \(x > 0\))
\(\eqalign{ & \Leftrightarrow {{\sqrt x - 1} \over {\sqrt x }} < 0\cr& \Leftrightarrow \sqrt x - 1 < 0\,\,\,\left( {\text{Vì }\,\sqrt x > 0\,khi\,x > 0} \right) \cr & \Leftrightarrow \sqrt x < 1 \Leftrightarrow 0 < x < 1 \cr} \)
Bài 5. Điều kiện : \(x ≥ 0\).
Ta có:
\(\eqalign{ & \left( {3 - 2\sqrt x } \right)\left( {2 + 3\sqrt x } \right) = 16 - 6x \cr & \Leftrightarrow 6 + 9\sqrt x - 4\sqrt x - 6x = 16 - 6x \cr & \Leftrightarrow 5\sqrt x = 10 \cr & \Leftrightarrow \sqrt x = 2 \cr} \)
\(\;\;⇔ x = 4\) (thỏa mãn điều kiện)
