Rút gọn \(K = \sqrt {\sqrt 5 {\rm{\;}} - \sqrt {3 - \sqrt {29 - 12\sqrt 5 } } } \)
Rút gọn \(K = \sqrt {\sqrt 5 {\rm{\;}} - \sqrt {3 - \sqrt {29 - 12\sqrt 5 } } } \)
Áp dụng hằng đẳng thức \(\sqrt {{A^2}} {\rm{\;}} = \left| A \right| = \left\{ {\begin{array}{*{20}{l}}{A{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} khi{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} A \ge 0}\\{ - A{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} khi{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} A < 0}\end{array}} \right..\)
Ta có:
\(K = \sqrt {\sqrt 5 {\rm{\;}} - \sqrt {3 - \sqrt {29 - 12\sqrt 5 } } } \)
\(\; = \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {20 - 2.3.2\sqrt 5 + 9} } } \)
\(\; = \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {{{(2\sqrt 5 - 3)}^2}} } } \)
\(\; = \sqrt {\sqrt 5 - \sqrt {3 - \left| {2\sqrt 5 - 3} \right|} } \)
\(\; = \sqrt {\sqrt 5 - \sqrt {3 - (2\sqrt 5 - 3)} } \) \(({\rm{do\;}}\,\,2\sqrt 5 - 3 > 0)\)
\(\; = \sqrt {\sqrt 5 - \sqrt {6 - 2\sqrt 5 } } \)
\(\; = \sqrt {\sqrt 5 - \sqrt {5 - 2\sqrt 5 + 1} } \)
\(\; = \sqrt {\sqrt 5 - \sqrt {{{(\sqrt 5 - 1)}^2}} } \)
\(\; = \sqrt {\sqrt 5 - \sqrt {\left| {\sqrt 5 - 1} \right|} } \)
\(\; = \sqrt {\sqrt 5 - (\sqrt 5 - 1)} \) \(({\rm{do\;}}\,\,\sqrt 5 - 1 > 0)\)
\(\; = \sqrt 1 = 1\)
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