Bài 1. a. \(\sqrt {\left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)} \)\( = \sqrt {3 - 2} = 1\)
b. Ta có:
\(\eqalign{ & B > 0 \cr&\Rightarrow {B^2} = {\left( {\sqrt {4 + \sqrt 7 } + \sqrt {4 - \sqrt 7 } } \right)^2} \cr & = 4 + \sqrt 7 + 2\sqrt {\left( {4 + \sqrt 7 } \right)\left( {4 - \sqrt 7 } \right)} + 4 - \sqrt 7 \cr & = 8 + 2\sqrt {16 - 7} = 8 + 2.3 = 14 \cr & \Rightarrow B = \sqrt {14} \cr} \)
Bài 2. Ta có:
\(\eqalign{ & \sqrt {7 - 2\sqrt {10} } + \sqrt 2 \cr & = \sqrt {7 - 2.\sqrt 5 .\sqrt 2 } + \sqrt 2 \cr & = \sqrt {{{\left( {\sqrt 5 - \sqrt 2 } \right)}^2}} + \sqrt 2 \cr & = \left| {\sqrt 5 - \sqrt 2 } \right| + \sqrt 2 \cr & = \sqrt 5 - \sqrt 2 + \sqrt 2 = \sqrt 5 \cr} \)
Bài 3. Ta có:
\(\eqalign{ & \sqrt 2 + \sqrt 3 < \sqrt {10} \cr & \Leftrightarrow {\left( {\sqrt 2 + \sqrt 3 } \right)^2} < 10 \cr & \Leftrightarrow 5 + 2.\sqrt 2 .\sqrt 3 < 10 \cr & \Leftrightarrow 2\sqrt 6 < 5 \Leftrightarrow {\left( {2\sqrt 6 } \right)^2} < 25 \cr} \)
\(⇔ 24 < 25\) (luôn đúng).
Vậy \(\sqrt 2 + \sqrt 3 < \sqrt {10}\)
