Bài 2. Rút gọn các biểu thức:
\(M = \sqrt {3 - 2\sqrt 2 } - \sqrt {6 + 4\sqrt 2 } \)
\(N = \sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } \)
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\(\eqalign{
& M = \sqrt {3 - 2\sqrt 2 } - \sqrt {6 + 4\sqrt 2 }= \cr
& \sqrt {{{\left( {\sqrt 2 } \right)}^2} - 2\sqrt 2 .1 + {1^2}} - \sqrt {{{\left( 2 \right)}^2} + 2.2.\sqrt 2 + {{\left( {\sqrt 2 } \right)}^2}} \cr
& = \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} - \sqrt {{{\left( {2 + \sqrt 2 } \right)}^2}} \cr
& = \left| {\sqrt 2 - 1} \right| - \left| {2 + \sqrt 2 } \right| \cr
& = \sqrt 2 - 1 - 2 - \sqrt 2 = - 3 \cr} \)
\(\eqalign{
& N = \sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } \cr
& \Rightarrow {N^2} = {\left( {\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } } \right)^2} \cr
& = 2 + \sqrt 3 + 2\sqrt {\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)} + 2 - \sqrt 3 \cr
& = 4 + 2\sqrt {4 - 3} = 6 \cr} \)
Vì \(N > 0\) nên \(N^2 = 6 ⇒ N = \sqrt6\). Vậy \(N = \sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } = \sqrt 6 \).