Salut,
Punctul a): Din enunț avem că:
[tex]b^{lga}\cdot c^{lgb}\cdot a^{lgc}=c^{lga}\cdot a^{lgb}\cdot b^{lgc}.\ Logaritm\breve{a}m\ \hat{i}n\ baza\ 10: \\\\ lg(b^{lga}\cdot c^{lgb}\cdot a^{lgc})=lg(c^{lga}\cdot a^{lgb}\cdot b^{lgc})\Rightarrow\\\\\Rightarrow lgb^{lga}+lgc^{lgb}+lga^{lgc}=lgc^{lga}+lga^{lgb}+lgb^{lgc},\ de\ aici:\\\\lga\cdot lgb+lgb\cdot lgc+lgc\cdot lga=lga\cdot lgc+lgb\cdot lga+lgc\cdot lgb,\ adev\breve{a}rat.[/tex]
Punctul b):
Știm că:
[tex]log_a\sqrt[p]x^k=log_ax^{\frac{k}{p}}=\dfrac{k}p\cdot log_ax.\ Num\breve{a}r\breve{a}torul\ frac\c{t}iei\ devine:\\\\\dfrac{1}n\cdot log_ax+\dfrac{3}n\cdot log_ax+\ldots+\dfrac{2n-1}n\cdot log_ax=\dfrac{1+3+\ldots+2n-1}n\cdot log_ax.\\\\\sum\limits_{k=1}^n(2k-1)=2\cdot\sum\limits_{k=1}^nk-\sum\limits_{k=1}^n1=2\cdot\dfrac{n(n+1)}2-n=n^2+n-2=n^2.\\\\Num\breve{a}r\breve{a}torul\ frac\c{t}iei\ devine:\ \dfrac{n^2}n\cdot log_ax=n\cdot log_ax.\ (1)\\\ Numitorul\ frac\c{t}iei\ devine:\\\\\dfrac{2}n\cdot log_ax+\dfrac{4}n\cdot log_ax+\ldots+\dfrac{2n}n\cdot log_ax=\dfrac{2+4+\ldots+2n}n\cdot log_ax=\\\\=\dfrac{2(1+2+\ldots+n)}n\cdot log_ax=\dfrac{2\cdot\dfrac{n(n+1)}2}n\cdot log_ax=(n+1)\cdot log_ax\ (2).[/tex]
Din relațiile (1) și (2), dacă le împarți, obții exact relația din enunț.
Green eyes.
