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[QUANVU]

Chứng minh rằng với mỗi số nguyên dương http://dientuvietnam...n/mimetex.cgi?n và mỗi các số thực dương http://dientuvietnam...metex.cgi?a,b,c ta có

[kimtruyen]

[quote name='QUANVU' date='September 11, 2006 05:33 pm'] Chứng minh rằng với mỗi số nguyên dương http://dientuvietnam...n/mimetex.cgi?n và mỗi các số thực dương http://dientuvietnam...metex.cgi?a,b,c ta có http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{a^{n+1}}{a^{n}+a^{n-1}b+\ldots+b^{n}}+\dfrac{b^{n+1}}{b^{n}+b^{n-1}c+\ldots+c^{n}}+\dfrac{c^{n+1}}{c^{n}+c^{n-1}a+\ldots+a^{n}}
B=http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{b^{n+1}}{a^{n}+a^{n-1}b+\ldots+b^{n}}+\dfrac{c^{n+1}}{b^{n}+b^{n-1}c+\ldots+c^{n}}+\dfrac{a^{n+1}}{c^{n}+c^{n-1}a+\ldots+a^{n}}
do đó
http://dientuvietnam.net/cgi-bin/mimetex.cgi?A=http://dientuvietnam.net/cgi-bin/mimetex.cgi?=http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{a^{n+1}+b^{n+1}}{a^{n}+a^{n-1}b+\ldots+b^{n}}+\dfrac{b^{n+1}+c^{n+1}}{b^{n}+b^{n-1}c+\ldots+c^{n}}+\dfrac{b^{n+1}+a^{n+1}}{c^{n}+c^{n-1}a+\ldots+a^{n}}
mặt khác ta dễ có:
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{a^{n+1}+b^{n+1}}{a^{n}+a^{n-1}b+\ldots+b^{n}}+\dfrac{b^{n+1}+c^{n+1}}{b^{n}+b^{n-1}c+\ldots+c^{n}}+\dfrac{b^{n+1}+a^{n+1}}{c^{n}+c^{n-1}a+\ldots+a^{n}}
vậy ta có đpcm

Edited by kimtruyen, 11-09-2006 - 18:34.