b 
c http://dientuvietnam.net/cgi-bin/mimetex.cgi?\g{0} chứng minh :http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{a^2-b^2}{c}+\dfrac{b^2-c^2}{a}+\dfrac{c^2-a^2}{b}\geq3a-4b+c
#1
Posted 20-12-2005 - 07:29
dungCT
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Hạ sĩ
cho a b c http://dientuvietnam.net/cgi-bin/mimetex.cgi?\g{0} chứng minh :
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{a^2-b^2}{c}+\dfrac{b^2-c^2}{a}+\dfrac{c^2-a^2}{b}\geq3a-4b+c
b c http://dientuvietnam.net/cgi-bin/mimetex.cgi?\g{0} chứng minh :http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{a^2-b^2}{c}+\dfrac{b^2-c^2}{a}+\dfrac{c^2-a^2}{b}\geq3a-4b+c
PHONG TRUNG CHI THẦN
#2
Posted 24-12-2005 - 11:10
dungCT
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Hạ sĩ
không bác nào giải dùm à
PHONG TRUNG CHI THẦN
#3
Posted 02-02-2006 - 21:02
tanlsth
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Tiến Sĩ Diễn Đàn Toán
bài này bạn tách là ra thôi
Learn from yesterday,live for today,hope for tomorrow
The important thing is to not stop questioning
#4
Posted 02-02-2006 - 22:59
manocanh
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Sĩ quan
em giải cho anh DungCT :cho a
b
c http://dientuvietnam.net/cgi-bin/mimetex.cgi?\g{0} chứng minh :
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{a^2-b^2}{c}+\dfrac{b^2-c^2}{a}+\dfrac{c^2-a^2}{b}\geq3a-4b+c
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