Bài 1. a. Ta có:
\(\eqalign{ & A = 2 - \sqrt {6 - 2\sqrt 5 } \cr& = 2 - \sqrt {{{\left( {1 - \sqrt 5 } \right)}^2}} = 2 - \left| {1 - \sqrt 5 } \right| \cr & = 2 + \left( {1 - \sqrt 5 } \right) = 3 - \sqrt 5 \cr} \)
(Vì \( 1 - \sqrt 5 < 0 \Rightarrow \left| {1 - \sqrt 5 } \right| \)\(= - \left( {1 - \sqrt 5 } \right)\) )
b. Ta có:
\(\eqalign{ & B = \left( {\sqrt {10} + \sqrt 6 } \right)\sqrt {{{\left( {\sqrt 3 - \sqrt 5 } \right)}^2}} \cr&= \sqrt 2 \left( {\sqrt 5 + \sqrt 3 } \right)\left| {\sqrt 3 - \sqrt 5 } \right| \cr & = \sqrt 2 \left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right) = 2\sqrt 2 \cr} \)
(Vì \(\sqrt 3 - \sqrt 5 < 0\, \)\(\Rightarrow \left| {\sqrt 3 - \sqrt 5 } \right| = - \left( {\sqrt 3 - \sqrt 5 } \right) \)\(= \sqrt 5 - \sqrt 3\) )
Bài 2. Ta có:
\(\eqalign{ & \sqrt {xy} + 2\sqrt x - 3\sqrt y - 6 \cr & = \sqrt x \left( {\sqrt y + 2} \right) - 3\left( {\sqrt y + 2} \right) \cr & = \left( {\sqrt y + 2} \right)\left( {\sqrt x - 3} \right) \cr} \)
Bài 3. Ta có:
\(\eqalign{ & \sqrt x + \sqrt {1 - x} = 1 \cr& \Leftrightarrow \left\{ {\matrix{ {x \ge 0} \cr {1 - x \ge 0} \cr {{{\left( {\sqrt x + \sqrt {1 - x} } \right)}^2} = 1} \cr } } \right. \cr & \Leftrightarrow \left\{ {\matrix{ {0 \le x \le 1} \cr {x + 2\sqrt {x\left( {1 - x} \right)} + 1 - x = 1} \cr } } \right.\cr& \Leftrightarrow \left\{ {\matrix{ {0 \le x \le 1} \cr {\sqrt {x\left( {1 - x} \right)} = 0} \cr } } \right. \cr & \Leftrightarrow \left[ {\matrix{ {x = 0} \cr {x = 1} \cr } } \right. \cr} \)
