$\lceil$ https://diendantoanh...dt/#entry719948 $\rfloor$
$\lceil$ To*Arthur*Pendragon $\it{!}$ $\rfloor$
$\sum\,\frac{\it{3}}{\it{5}- \it{6}\,\it{ab}}=$ $\sum\,\frac{\it{3}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}}{\it{5}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}- \it{6}\,\it{ab}}\leqq$ $\sum\,\frac{\it{6}\,\it{c}^{\,\it{2}}+ \it{12}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}\,\,\it{)}+ \it{(}\,\,\sqrt{\it{629}}- \it{5}\,\,\it{)}\it{(}\,\,\it{bc}+ \it{ca}\,\,\it{)}+ \it{(}\,\,\it{91}- \it{2}\sqrt{\it{629}}\,\,\it{)}\it{ab}}{\it{27}\it{(}\,\,\it{ab}+ \it{bc}+ \it{ca}\,\,\it{)}+ \it{10}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}}=$ $\it{3}$
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