Cho dãy {$x_{n} $}được xác định như sau:
$x_{1} \geq \dfrac{3}{4}; x_{n+1}^{4}+3 =4x_{n}$.CMR
$2x_{n+5}^{4}+3 \leq x_{n-5}^{4} +4 \sqrt[4]{4x_{n-5}-3} $
Edited by Oral1020, 08-06-2013 - 10:54.
$2x_{n+5}^{4}+3 \leq x_{n-5}^{4} +4 \sqrt[4]{4x_{n-5}-3} $
Started By thanhbinh214, 07-11-2007 - 23:21
#1
Posted 07-11-2007 - 23:21
thanhbinh214
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#2
Posted 08-06-2013 - 09:47
voidaudan
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Lính mới
<=> $ 2.x_{n+4} \le x_{n-6}+x_{n-4} $
do $ x_{n} \ge x_{n+1} $ => DPCM
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